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whai 2024-03-22 22:57:40 +08:00
parent 99bda3a447
commit a2d5497bae
27 changed files with 1690 additions and 14 deletions
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4
pom.xml
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<version>1.0-SNAPSHOT</version> <version>1.0-SNAPSHOT</version>
<properties> <properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<maven.compiler.source>1.8</maven.compiler.source> <maven.compiler.source>21</maven.compiler.source>
<maven.compiler.target>1.8</maven.compiler.target> <maven.compiler.target>21</maven.compiler.target>
</properties> </properties>

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src/main/java/cn/whaifree/LCR/LCR001.java Normal file
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package cn.whaifree.LCR;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 11:35
* @注释
*/
public class LCR001 {
@Test
public void test() {
int a = 15;
int b = 2;
int c = new Solution1().divide(a, b);
System.out.println(c);
}
class Solution {
/**
* 位运算
* @param a
* @param b
* @return
*/
public int divide(int a, int b) {
/**
* 17/3
* 3<2 2次 一直到 24-12 ，之间有2^2=4个3，剩下的5再 3<2 只要0次，则5中有2^0个3
*/
boolean reverse = false;
if ((a < 0 && b > 0) || (a > 0 && b < 0)) {
reverse = true;
}
if (a < 0) {
a = -a;
}
if (b < 0) {
b = -b;
}
int down = 0;
int sum = 0;
int tmp = a;
while (tmp > 0) {
sum += down;
tmp -= down * b;
down = getDown(tmp, b);
}
return reverse ? -sum : sum;
}
/**
*
* @param beDivide
* @param divide
* @return beDivide包含几个divide
*/
public int getDown(int beDivide, int divide) {
int have = 0;
int tmp = divide;
while (tmp < beDivide) {
have++;
tmp = tmp << 2;
}
return 1 << have;
}
}
class Solution1 {
public int divide(int a, int b) {
//特殊情况1, b=1
if (b == 1){
return a;
}
//特殊情况2, b=-1
if (b == -1){
return a == Integer.MIN_VALUE ? Integer.MAX_VALUE : -a;
}
//特殊情况3, a=0
if (a == 0){
return 0;
}
//确定符号
boolean positive = (a ^ b) >= 0;
//为避免溢出, 转换为负数进行计算
a = a < 0 ? a : -a;
b = b < 0 ? b : -b;
//快速相减
/**
* 17/3
* 3<2 2次 一直到 24-12 ，之间有2^2=4个3，剩下的5再 3<2 只要0次，则5中有2^0个3
*/
int quotient = 0;
while (a <= b){
int base = 1;
int divisor = b;
//使用减法, 避免溢出
while (a - divisor <= divisor){
divisor <<= 1; // 被除数*2 b=2时 就是 2 4 8 16 不断扩大，直到达到a的一半
base <<= 1; // 表示包含的（b 被除数）的个数，后面给他加到quotient中
}
quotient += base;
a -= divisor; // a减去前面已经找到的部分 即 17 找到了前面的12，就17-12=5
}
return positive ? quotient : -quotient;
}
}
class Solution2 {
public int divide(int a, int b) {
// 不考虑越界的写法
int res = 0;
while (a < b) {
int count = 1;//记住有几个要加入
int copyB = b; // 记录每次的总量
while (a - copyB < copyB) {
copyB <<= 1;
count <<= 1;
}
res += count;
a -= copyB;
}
return res;
}
}
}

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src/main/java/cn/whaifree/LCR/LCR089.java Normal file
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package cn.whaifree.LCR;
import cn.whaifree.interview.Meituan.Test02;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 20:05
* @注释
*/
public class LCR089 {
@Test
public void test() {
int[] values = {2, 7, 9, 3, 1};
int result = new Solution().rob(values);
System.out.println(result);
// 类似题目
new Test02().sugar(values);
}
/**
*
*/
class Solution {
public int rob(int[] values) {
if (values.length == 1) {
return values[0];
}
// dp[i]表示当前最大价值
// dp[i]=max(dp[i-1],dp[i-2]+values[i])
/**
*
* 2,7,9,3,1
* 2 7 11 11 12
*
*/
// 初始化前3个
int[] dp = new int[values.length];
dp[0] = values[0];
dp[1] = Math.max(values[0], values[1]);
for (int i = 2; i < dp.length; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + values[i]);
}
return dp[values.length - 1];
}
}
}

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src/main/java/cn/whaifree/LCR/LCR090.java Normal file
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package cn.whaifree.LCR;
import org.junit.Test;
import java.util.HashMap;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 20:16
* @注释
*/
public class LCR090{
@Test
public void test()
{
HashMap<Object, Object> objectObjectHashMap = new HashMap<>();
objectObjectHashMap.put(null, "");
System.out.println(objectObjectHashMap.get(null));
System.out.println(objectObjectHashMap.get("a"));
int[] nums = {1,2,1,1};
System.out.println(new Solution().rob(nums));
}
class Solution {
public int rob(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
return Math.max(nums[0], nums[1]);
}
return Math.max(
robRange(nums, 0, nums.length - 2),
robRange(nums, 1, nums.length - 1)
);
}
public int robRange(int[] nums, int start, int end) {
int first = nums[start], second = Math.max(nums[start], nums[start + 1]);
for (int i = start + 2; i <= end; i++) {
int temp = second;
second = Math.max(first + nums[i], second);
first = temp;
}
return second;
}
}
}

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src/main/java/cn/whaifree/interview/Meituan/Test01.java Normal file
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package cn.whaifree.interview.Meituan;
import org.junit.Test;
import java.util.Scanner;
/**
*
* <a href="https://www.yuque.com/taxing-qarr6/hxitgt/acebypk1uc68zhi2">...</a>
*
* @version 1.0
* @Author whai文海
* @Date 2024/3/19 11:10
* @注释
*/
public class Test01 {
@Test
public void test() {
String s = "CSDFMTMTDSA";
int k = 2;
System.out.println(mt(k, s));
}
/**
* * MT 是美团的缩写，因此小美很喜欢这两个字母。
* * 现在小美拿到了一个仅由大写字母组成字符串，她可以最多操作k次，每次可以修改任意一个字符。小美想知道，操作结束后最多共有多少个"M'和'T'字符?
* * 输入描述
* * 第一行输入两个正整数n,k，代表字符串长度和操作次数。第二行输入一个长度为n的、仅由大写字母组成的字符串。1<=k<=n<=10^5
* * 输出描述
* * 输出操作结束后最多共有多少个'M'和'T'字符。
* * 示例 1
* * 输入
* * C++
* * 5 2
* * M鉀锯撂駙拳顰棛薪廈阗摟粒锟惠拱➋壢草麦耸蟭骚殝騶乳
* * 输出
* * Plain Text
* * A
* * A
* * 说明
* * 修改第三个和第五个字符，形成的字符串为 MTTAM，这样共有 4 个'M'和'T'。
*
* @param k
* @param s
* @return
*/
public int mt(int k, String s) {
int isMt = 0;
char[] chars = s.toCharArray();
for (char aChar : chars) {
if (aChar == 'M' || aChar == 'T') {
isMt++;
}
}
int isNotMt = chars.length - isMt;
if (isNotMt > k) {
return k + isMt;
}else{
return chars.length;
}
}
public static void main(String[] args) {
new Test01().mt2();
// new Test01().mt4();
}
/**
* 小美拿到了一个由正整数组成的数组，但其中有一些元素是未知的(用 0 来表示)。
* 现在小美想知道，如果那些未知的元素在区间[,]范围内随机取值的话，数组所有元素之和的最小值和最大值分别是多少?
* 共有q次询问。
* 输入描述
* 第一行输入两个正整数n,q，代表数组大小和询问次数。
* 第二行输入n个整数ai，其中如果输入ai的为 0，那么说明ai是未知的。
* 接下来的q行，每行输入两个正整数1r，代表一次询问。
* Plain Text
* 1<=n,q<=10^5
* 0<=ai<=10^9
* 31<=l<=r<=10^9
* 输出描述
* 输出q行，每行输出两个正整数，代表所有元素之和的最小值和最大值。
*/
public void mt2() {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int query = scanner.nextInt();
int[] nums = new int[n];
int sum = 0;
int isZero = 0;
for (int i = 0; i < nums.length; i++) {
int inputNumber = scanner.nextInt();
if (inputNumber==0) isZero++;
nums[i] = inputNumber;
sum += inputNumber;
}
for (int i = 0; i < query; i++) {
int left = scanner.nextInt();
int right = scanner.nextInt();
System.out.println("=====");
System.out.print(sum + left * isZero + " " + (sum + right * isZero));
}
}
public void mt4(int[] nums, int k) {
int sum = 1;
for (int num : nums) {
sum *= num;
}
int ans = 0;
for (int i = 0; i < nums.length; i++) {
int sub = sum / nums[i];
if (nums.length - String.valueOf(sub).lastIndexOf("0") == k) {
ans++;
}
}
System.out.println(ans);
}
}

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src/main/java/cn/whaifree/interview/Meituan/Test02.java Normal file
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package cn.whaifree.interview.Meituan;
import org.junit.Assert;
import org.junit.Test;
import java.util.*;
/**
* <a href="https://www.nowcoder.com/discuss/599738882497736704?sourceSSR=search">...</a>
*
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 19:14
* @注释
*/
public class Test02 {
public static void main(String[] args) {
new Test02().test();
}
public void test() {
Scanner scanner = new Scanner(System.in);
int number = scanner.nextInt();
for (int i = 0; i < number; i++) {
int num = scanner.nextInt();
int[] input = new int[num];
for (int j = 0; j < num; j++) {
input[j] = scanner.nextInt();
}
int[] output = new int[num];
for (int j = 0; j < num; j++) {
output[j] = scanner.nextInt();
}
isSuccess(input, output);
}
}
public void isSuccess(int[] input, int[] output) {
LinkedList<Integer> stack = new LinkedList<>();
int index = 0;
for (int i : input) {
stack.push(i);
while (!stack.isEmpty() && stack.peek() == output[index++]) {
stack.pop();
}
}
if (stack.isEmpty()) {
System.out.println("yes");
} else {
System.out.println("no");
}
}
@Test
public void test1() {
int[] input = {1, 2, 3, 4, 5};
int[] output = {4, 5, 3, 2, 1};
System.out.println(new Solution946().validateStackSequences(input, output));
}
class Solution946 {
public boolean validateStackSequences(int[] pushed, int[] popped) {
return success(pushed, popped);
}
public boolean success(int[] input, int[] output) {
LinkedList<Integer> stack = new LinkedList<>();
int index = 0;
for (int i : input) {
stack.push(i);
while (!stack.isEmpty() && stack.peek() == output[index]) {
index++;
stack.pop();
}
}
return stack.isEmpty();
}
}
@Test
public void test2() {
sugar(new int[]{3, 1, 2, 7, 10, 2, 4});
}
/**
* @param values
*/
public void sugar(int[] values) {
// dp[i]表示当前最大价值
// dp[i]=max(dp[i-1],dp[i-3]+values[i])
/**
*
* 3 1 2 7 10 2 4
*
* 3 3 3 10 13 13 14
*
*/
// 初始化前3个
int[] dp = new int[values.length];
int max = 0;
for (int i = 0; i < 3; i++) {
max = Math.max(max, values[i]);
}
for (int i = 0; i < 3; i++) {
dp[i] = max;
}
for (int i = 3; i < dp.length; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 3] + values[i]);
}
System.out.println(Arrays.toString(dp));
}
/**
* dp[i]表示从商品0-i中的最美味值
* <p>
* 获取最大的k个值，遇到最大的这k个值就必吃
* <p>
* dp[i] =
* 1. 如果values[i]不存在于k大个值，如果这个不吃 dp[i-1]
* 2. 如果吃 dp[i-2] + values[i]
* 3. 如果存在于最大的k个值 dp[i-1] + values[i]
* <p>
* 1 2 3 4 5 6 7
* <p>
* 0 1 2 3 4 5 6
* 1 2 4 6 9 12 16
*
* @param values 美味值
* @param k 允许打破的次数
*/
public int calculateMaxTaste(int[] values, int k) {
PriorityQueue<Integer> priorityQueue = new PriorityQueue<>(k, Comparator.reverseOrder());
for (int value : values) {
priorityQueue.add(value);
}
ArrayList<Integer> list = new ArrayList<>();
for (int i = 0; i < k; i++) {
list.add(priorityQueue.poll());
}
int[] dp = new int[values.length];
dp[0] = values[0];
dp[1] = Math.max(values[0], values[1]);
for (int i = 2; i < dp.length; i++) {
if (list.contains(values[i])) {
System.out.println(i + "打破规则:" + values[i]);
dp[i] = dp[i - 1] + values[i]; //打破规则
} else {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + values[i]);
}
}
return dp[values.length - 1];
}
@Test
public void test3() {
int[] values = {1, 2, 3, 4, 5, 6, 7};
int k = 1;
// sugar(values, k);
}
@Test
public void testCalculateMaxTaste() {
Test02 strategy = new Test02();
// 测试用例1
int n1 = 3, k1 = 1;
int[] delicious1 = {1, 2, 3};
Assert.assertEquals(5, strategy.calculateMaxTaste(delicious1, k1));
// 测试用例2
int n2 = 4, k2 = 2;
int[] delicious2 = {1, 2, 3, 4};
Assert.assertEquals(9, strategy.calculateMaxTaste(delicious2, k2));
// 测试用例3
int n3 = 5, k3 = 1;
int[] delicious3 = {1, 3, 2, 4, 5};
Assert.assertEquals(12, strategy.calculateMaxTaste(delicious3, k3));
// 测试用例4
int n4 = 3, k4 = 0;
int[] delicious4 = {3, 1, 2};
Assert.assertEquals(3, strategy.calculateMaxTaste(delicious4, k4));
// 测试用例5
int n5 = 4, k5 = 1;
int[] delicious5 = {1, 1, 1, 1};
Assert.assertEquals(3, strategy.calculateMaxTaste(delicious5, k5));
// 测试用例6
int n6 = 6, k6 = 2;
int[] delicious6 = {5, 2, 4, 1, 3, 6};
Assert.assertEquals(16, strategy.calculateMaxTaste(delicious6, k6));
// 测试用例7
int n7 = 5, k7 = 2;
int[] delicious7 = {2, 3, 1, 5, 4};
Assert.assertEquals(11, strategy.calculateMaxTaste(delicious7, k7));
// 测试用例8
int n8 = 4, k8 = 0;
int[] delicious8 = {4, 3, 2, 1};
Assert.assertEquals(4, strategy.calculateMaxTaste(delicious8, k8));
// 测试用例9
int n9 = 6, k9 = 3;
int[] delicious9 = {2, 1, 5, 3, 6, 4};
Assert.assertEquals(19, strategy.calculateMaxTaste(delicious9, k9));
// 测试用例10
int n10 = 7, k10 = 1;
int[] delicious10 = {1, 2, 3, 4, 5, 6, 7};
Assert.assertEquals(26, strategy.calculateMaxTaste(delicious10, k10));
}
}

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src/main/java/cn/whaifree/leetCode/Array/LeetCode912_SortArrays.java Normal file
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package cn.whaifree.leetCode.Array;
import org.junit.Test;
import java.util.Arrays;
/**
*
* 排序算法有这几种：
* 常见的排序算法包括但不限于以下这些：
* 冒泡排序：从第一个元素开始与右侧元素两两比较并交换，直到右侧成为有序部分。
* 选择排序：有序部分在左侧，在剩余元素中找到最小的那个元素，并与剩余元素中第一个元素交换。
* 插入排序：有序部分在左侧，将剩余元素中第一个元素不断向左交换，直到此元素处于有序部分恰当位置。
* 希尔排序：取一个间隔值，距离为间隔值的元素为一组，将整个数组分为若干组。每组内进行插入排序。缩小间隔值并重复，直到间隔值为1，即所有元素在同一组。
*
* @version 1.0
* @Author whai文海
* @Date 2024/3/20 14:20
* @注释
*/
public class LeetCode912_SortArrays {
@Test
public void test() {
int[] nums = {5, 2, 3, 1,647,24,7,2,8,2,8,1,54,13,6,234,45,234,64,745,32,56,44,32,38};
int[] res = new Solution2().sortArray(nums);
System.out.println(Arrays.toString(res));
int[] ints = Arrays.copyOf(res, res.length);
Arrays.sort(nums);
System.out.println(Arrays.equals(ints, nums));
}
class Solution {
/**
* 冒泡
* @param nums
* @return
*/
public int[] sortArray(int[] nums) {
boolean isSwap = false;
for (int i = 0; i < nums.length; i++) {
for (int j = 1; j < nums.length - i; j++) {
if (nums[j] < nums[j - 1]) {
isSwap = true;
int tmp = nums[j];
nums[j] = nums[j - 1];
nums[j - 1] = tmp;
}
}
if (!isSwap) {
return nums;
}
isSwap = false;
}
return nums;
}
}
class Solution1 {
/**
* 选择排序
* 选择右边区间的最小值与左边替换
* @param nums
* @return
*/
public int[] sortArray(int[] nums) {
for (int i = 0; i < nums.length; i++) {
int minIndex = i;
for (int j = i+1; j < nums.length; j++) {
if (nums[minIndex] > nums[j]) {
minIndex = j;
}
}
// 交换min与i
int tmp = nums[i];
nums[i] = nums[minIndex];
nums[minIndex] = tmp;
}
return nums;
}
}
class Solution2 {
/**
* 插入排序 j不断往前插入替换到合适位置
* @param nums
* @return
*/
public int[] sortArray(int[] nums) {
for (int i = 1; i < nums.length; i++) {
int j = i;
while (j > 0 && nums[j] < nums[j - 1]) {
// 交换j和j-1
int tmp = nums[j];
nums[j] = nums[j - 1];
nums[j - 1] = tmp;
j--;
}
}
return nums;
}
}
class SolutionQuickSort {
/**
* 插入排序 j不断往前插入替换到合适位置
*
*
* @param nums
* @return
*/
public int[] sortArray(int[] nums) {
return nums;
}
public void part(int[] nums, int start, int end) {
if (start > end) {
return;
}
}
}
}

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src/main/java/cn/whaifree/leetCode/Dynamic/LeetCode1049.java Normal file
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package cn.whaifree.leetCode.Dynamic;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/18 14:31
* @注释
*/
public class LeetCode1049 {
@Test
public void test() {
int[] stones = {1,2};
int result = new Solution1().lastStoneWeightII(stones);
System.out.println(result);
}
class Solution {
/**
* 背包容量为half，让其尽可能装满，除非装不下了。到装不下了，这一部分已经装入的就是拿来碰撞的。剩下就是sum-dp[half]-dp[half]
*
* 目标就是让石头的总价值分为两部分
*
* 转换为背包问题
*
* dp[j] 表示 容量为j的背包最多能够容纳多少价值的物品
* - 物品i的价值 为 stones[i]
* - 物品i的重量 为 stones[i]
*
* 1 <= stones.length <= 30
* 1 <= stones[i] <= 100
* 初始化为store[half]
*
*
* 递推公式 dp[j] = Math.max(dp[j],dp[j-weight[i]]+value[i])
* 分成两堆石头，一堆石头的总重量是dp[half]，另一堆就是sum - dp[half]。
*
* @param stones
* @return
*/
public int lastStoneWeightII(int[] stones) {
int sum = 0;
for (int stone : stones) {
sum += stone;
}
int half = sum / 2;
int[] dp = new int[half + 1];
for (int i = 0; i < stones.length; i++) {
for (int j = half; j >= stones[i]; j--) { // 确保背包能够放下
dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
}
}
// dp[half] 是要被消灭的，那两个部分的消灭就*2
return sum - dp[half] * 2;
}
}
class Solution1 {
/**
* half = sum/2
*
*
* dp[i][j] 表示 放入i后，容量为j的背包的最大价值
*
* 1. 放入i dp[i][j] = dp[i-1][j-weight[i]]+value[i]
* 2. 不放入i dp[i][j] = dp[i-1][j]
*
* 2 4 1 1 target=4
*
* 0 1 2 3 4
* 0 0 0 2 2 2
* 1 0 0 2 2 2
* 2 0 1 3 3 3
* 3 0 1 4 4 4
*
* @param stones
* @return
*/
public int lastStoneWeightII(int[] stones) {
int sum = 0;
for (int stone : stones) {
sum += stone;
}
// 计算所有石头的总重量，并将其的一半作为目标重量
int half = sum / 2;
int length = stones.length;
int[][] dp = new int[length][half + 1];
for (int i = stones[0]; i <= half; i++) {
dp[0][i] = stones[0];
}
for (int i = 1; i < length; i++) {
for (int j = 1; j <= half ; j++) {
if (j < stones[i]) {
// 放不下 就不放了，价值为i-1物品的最大价值
// 当前石头重量大于目标重量，无法放入，价值等于前i-1个石头的最大价值
dp[i][j] = dp[i - 1][j];
}else {
// 取放入和不放入当前石头两种情况中的最大价值
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - stones[i]] + stones[i]);
}
}
}
// 返回石头总重量减去最接近目标重量的那堆石头的两倍重量
return sum - dp[length - 1][half] * 2;
}
}
}

173
src/main/java/cn/whaifree/leetCode/Dynamic/LeetCode494.java Normal file
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@ -0,0 +1,173 @@
package cn.whaifree.leetCode.Dynamic;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/18 16:17
* @注释
*/
public class LeetCode494 {
@Test
public void test()
{
int[] nums = {1,1,1,1,1};
int target = 3;
System.out.println(new Solution2().findTargetSumWays(nums, target));
}
class Solution {
/**
* 背包容量为3
* left（+1的数量）-right（-1的数量） = target
* left-right = target
* left+right = sum （right = sum -left）
* left = (target + sum)/2
* 表示正数有多少个是固定的，就是我们的背包容量
*
*
* 使用nums[i] 里面全是1 装满容量为left的背包，有几种方法
*
* if j> nums[i]
* dp[i][j] = dp[i-1][j] + dp[i-1][j-nums[i]]
* 不放 放
* else
* dp[i][j] = dp[i-1][j]
* 1,1,1,1,1
*
* 0 1 2 3 4
* 0 1 1 0 0 0
* 1 0 0 1 0 0
* 2 0 3
* 3
* 4
*
* @param nums
* @param target
* @return
*/
public int findTargetSumWays(int[] nums, int target) {
int sum = 0;
for (int num : nums) {
sum += num;
}
// 总和还比不过绝对值，1不够用
if(sum < Math.abs(target)){
return 0;
}
// left 为+1的数量 +1的数量必须是整数
if((sum + target) % 2 != 0) {
return 0;
}
int left = (sum + target) >> 1; // (sum + target) / 2;
int length = nums.length;
int[][] dp = new int[length + 1][left + 1];
// 01背包
// i(1 ~ len)表示遍历（不一定选）了 i 个元素，j(0 ~ sum) 表示它们的和
dp[0][0] = 1;
for (int i = 1; i <= length; i++) {
for (int j = 0; j <= left; j++) {
// 装不下（不选当前元素）
if (j - nums[i - 1] < 0) {
dp[i][j] = dp[i - 1][j];
// 可装可不装（当前元素可选可不选）
} else {
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - nums[i - 1]];
}
}
}
/**
* 初始化：0个元素，和为0，情况有1种（因为没有元素，所以只能不选，和为0）：dp[0][0] = 1
* 不选当前元素，即"部分和"(即j)与之前相同：dp[i][j] = dp[i - 1][j]
* 可选可不选，不选的情况是2，选当前元素的话则之前的状态应为dp[i - 1][j - num]（这里的num指的是当前元素的值，即代码中的nums[i - 1]），二者相加，即：dp[i][j] = dp[i - 1][j] + dp[i - 1][j - num]
*/
return dp[length][left];
}
}
class Solution1 {
public int findTargetSumWays(int[] nums, int S) {
int sum = 0;
for (int num : nums) {
sum += num;
}
// 背包容量为整数，sum + S为奇数的话不满足要求
if (((sum + S) & 1) == 1) {
return 0;
}
// 目标和不可能大于总和
if (S > sum) {
return 0;
}
sum = (sum + S) >> 1;
int len = nums.length;
int[][] dp = new int[len + 1][sum + 1];
dp[0][0] = 1;
// 如果迭代部分 j 的初值赋 1 的话，就要先初始化 j = 0 的情况
/* int count = 1;
for (int i = 1; i <= len; i++) {
// ±0 均可
if (nums[i - 1] == 0) {
count *= 2;
}
dp[i][0] = count;
} */
// 01背包
// i(1 ~ len)表示遍历（不一定选）了 i 个元素，j(0 ~ sum) 表示它们的和
for (int i = 1; i <= len; i++) {
for (int j = 0; j <= sum; j++) {
// 装不下（不选当前元素）
if (j - nums[i - 1] < 0) {
dp[i][j] = dp[i - 1][j];
// 可装可不装（当前元素可选可不选）
} else {
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - nums[i - 1]];
}
}
}
return dp[len][sum];
}
}
class Solution2 {
public int findTargetSumWays(int[] nums, int S) {
int sum = 0;
for(int i=0; i<nums.length; i++){
sum += nums[i];
nums[i] += nums[i];
}
if(S>sum) return 0;
S += sum;
int[] dp = new int[S+1];
dp[0] = 1;
for(int num: nums){
for(int i=S; i>=0; i--){
if(i-num>=0){
dp[i] = dp[i] + dp[i-num];
}
}
}
return dp[S];
}
}
}

98
src/main/java/cn/whaifree/leetCode/Dynamic/LeetCode647.java Normal file
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@ -0,0 +1,98 @@
package cn.whaifree.leetCode.Dynamic;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/22 12:36
* @注释
*/
public class LeetCode647 {
class Solution {
/**
* 判断一个子字符串（字符串的下表范围[i,j]）是否回文，
* 依赖于子字符串（下表范围[i + 1, j - 1]）） 是否是回文。
*
* 布尔类型的dp[i][j]：表示区间范围[i,j] （注意是左闭右闭）的子串是否是回文子串
*
* ij两个循环，能够覆盖所有循环
* - dp[i][j] 判断是否回文的时候，只需要dp[i+1][j-1]是否回文，并且s[i]==s[j]
* 当s[i]=s[j]
* 1. i=j 回文
* 2. i+1=j 回文
* 3. dp[i+1][j-1]==true 回文
*
*
* <img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210121171032473-20230310132134822.jpg" alt="">
* dp[i][j]由dp[i+1][j-1]过来，所以必须从下到上，左到右
*
* @param s
* @return
*/
public int countSubstrings(String s) {
char[] chars = s.toCharArray();
boolean[][] dp = new boolean[chars.length][chars.length];
int res = 0;
for (int i = s.length() - 1; i >= 0; i--) {
for (int j = i; j < s.length(); j++) {
if (chars[i] == chars[j]) {
if (j - i == 1|| i==j || dp[i + 1][j - 1]) {
dp[i][j] = true;
res++;
}
}
}
}
return res;
}
}
class Solution1 {
public int countSubstrings(String s) {
// 中心点向外扩充
// 1个元素可以作为中心点
// 2个元素也能作为中心点
char[] chars = s.toCharArray();
int res = 0;
for (int i = 0; i < chars.length; i++) {
res += subString(chars, i, i);
res += subString(chars, i, i + 1);
}
return res;
}
/**
*
* @param chars
* @param center1 向左扩展的指针
* @param center2 向右扩展的指针
* @return 有几个子串
*/
public int subString(char[] chars, int center1, int center2) {
int res = 0;
while (center1 >= 0 && center2 < chars.length && chars[center1] == chars[center2]) {
center1--;
center2++;
res++;
}
return res;
}
}
@Test
public void test()
{
Solution1 solution = new Solution1();
String s = "aaa";
int res = solution.countSubstrings(s);
System.out.println(res);
}
}

2
src/main/java/cn/whaifree/leetCode/Greedy/LeetCode376.java
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@ -1,6 +1,6 @@
package cn.whaifree.leetCode.Greedy; package cn.whaifree.leetCode.Greedy;
import jdk.internal.instrumentation.InstrumentationTarget;
import org.junit.Test; import org.junit.Test;
/** /**

2
src/main/java/cn/whaifree/leetCode/Greedy/LeetCode55.java
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@ -2,7 +2,7 @@ package cn.whaifree.leetCode.Greedy;
import cn.whaifree.leetCode.Tree.LeetCode94; import cn.whaifree.leetCode.Tree.LeetCode94;
import org.junit.Test; import org.junit.Test;
import sun.misc.Launcher;
import java.net.URL; import java.net.URL;

95
src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode4.java Normal file
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@ -0,0 +1,95 @@
package cn.whaifree.leetCode.LeetCode;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 16:57
* @注释
*/
public class LeetCode4 {
@Test
public void test()
{
Solution2 solution = new Solution2();
int[] nums1 = {1,3};
int[] nums2 = {2};
double medianSortedArrays = solution.findMedianSortedArrays(nums1, nums2);
System.out.println(medianSortedArrays);
}
class Solution1 {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int length = nums1.length + nums2.length;
int middle = length / 2;
int index1 = 0;
int index2 = 0;
boolean flag = false;
while (index1 + index2 < middle - 1) {
if (nums1[index1] < nums2[index2]) {
flag = false;
index1++;
} else {
flag = true;
index2++;
}
}
if (length % 2 != 0) {
return flag ? nums2[index2] : nums1[index1];
}else {
return (double) (nums1[index1] + nums2[index2]) / 2;
}
}
}
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int[] nums = new int[nums1.length + nums2.length];
int indexA = 0;
int indexB = 0;
int index = 0;
while (indexA < nums1.length && indexB < nums2.length) {
if (nums1[indexA] <= nums2[indexB]) {
nums[index++] = nums1[indexA++];
}else {
nums[index++] = nums2[indexB++];
}
}
while (indexA < nums1.length) {
nums[index++] = nums1[indexA++];
}
while (indexB < nums2.length) {
nums[index++] = nums2[indexB++];
}
int length = nums.length;
return length % 2 == 1 ? (double) nums[length / 2] : (double) (nums[length/2 - 1] + nums[length/2]) / 2;
}
}
class Solution2 {
/**
* 二分查找
* @param nums1
* @param nums2
* @return
*/
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
return 1;
}
}
}

2
src/main/java/cn/whaifree/leetCode/LinkedList/LeetCode206.java
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@ -1,7 +1,7 @@
package cn.whaifree.leetCode.LinkedList; package cn.whaifree.leetCode.LinkedList;
import cn.whaifree.leetCode.model.ListNode; import cn.whaifree.leetCode.model.ListNode;
import com.sun.jmx.remote.internal.ArrayQueue;
import org.junit.Test; import org.junit.Test;
import java.util.List; import java.util.List;

37
src/main/java/cn/whaifree/leetCode/Stack/LeetCode946.java Normal file
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@ -0,0 +1,37 @@
package cn.whaifree.leetCode.Stack;
import org.junit.Test;
import java.util.LinkedList;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/21 19:38
* @注释
*/
public class LeetCode946 {
@Test
public void test() {
new Solution().validateStackSequences(new int[]{1, 2, 3, 4, 5}, new int[]{4, 5, 3, 2, 1});
}
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
return success(pushed, popped);
}
public boolean success(int[] input, int[] output) {
LinkedList<Integer> stack = new LinkedList<>();
int index = 0;
for (int i : input) {
stack.push(i);
while (!stack.isEmpty() && stack.peek() == output[index]) {
index++;
stack.pop();
}
}
return stack.isEmpty();
}
}
}

102
src/main/java/cn/whaifree/leetCode/String/LeetCode5.java Normal file
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@ -0,0 +1,102 @@
package cn.whaifree.leetCode.String;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/22 12:16
* @注释
*/
public class LeetCode5 {
@Test
public void test()
{
Solution1 solution = new Solution1();
String s = "cbbd";
String s1 = solution.longestPalindrome(s);
System.out.println(s1);
}
class Solution {
/**
* 动态规划
* dp[i][j] 表示 i-j的串是不是回文串
* 同时记录i和j的最大差值的i和j
*
* s[i] = s[j] 时
* dp[i][j] =
* 1. i=j true
* 2. i=j+1 true
* 3. if(dp[i+1][j-1]) true
*
*
* @param s
* @return
*/
public String longestPalindrome(String s) {
int length = s.length();
boolean[][] dp = new boolean[length][length];
int maxI = 0;
int maxJ = 0;
// dp[i+1][j-1] 决定了dp[i][j] 所以 要从下往上，从左往右遍历
for (int i = length; i >= 0; i--) {
for (int j = i; j < length; j++) {
if (s.charAt(i) == s.charAt(j) && (i == j || i + 1 == j || dp[i + 1][j - 1])) {
if ((maxJ - maxI) <= j - i) {
maxI = i;
maxJ = j;
}
dp[i][j] = true;
}
}
}
return s.substring(maxI, maxJ+1);
}
}
class Solution1 {
public String longestPalindrome(String s) {
char[] chars = s.toCharArray();
// 向外扩展，统计回文长度
int length = s.length();
int start = 0;
int end = 0;
for (int i = 0; i < length; i++) {
int a = getHuiWenLengthReturnLength(chars, i, i);
int b = getHuiWenLengthReturnLength(chars, i, i + 1);
int maxLength = Math.max(a, b);
if (maxLength > end - start){
start = i - (maxLength - 1) / 2;
end = i + maxLength / 2;
}
}
return s.substring(start, end + 1);
}
public String getHuiWenLength(char[] chars, int left, int right) {
while (left >= 0 && right < chars.length && chars[left] == chars[right]) {
left--;
right++;
}
return new String(chars, left + 1, right - left - 1);
}
public int getHuiWenLengthReturnLength(char[] chars, int left, int right) {
while (left >= 0 && right < chars.length && chars[left] == chars[right]) {
left--;
right++;
}
return right - left - 1;
}
}
}

2
src/main/java/cn/whaifree/leetCode/Tree/LeetCode107.java
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@ -1,7 +1,7 @@
package cn.whaifree.leetCode.Tree; package cn.whaifree.leetCode.Tree;
import cn.whaifree.leetCode.model.TreeNode; import cn.whaifree.leetCode.model.TreeNode;
import com.sun.org.apache.bcel.internal.generic.NEW;
import org.junit.Test; import org.junit.Test;
import java.util.*; import java.util.*;

2
src/main/java/cn/whaifree/leetCode/Tree/LeetCode404.java
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@ -1,7 +1,7 @@
package cn.whaifree.leetCode.Tree; package cn.whaifree.leetCode.Tree;
import cn.whaifree.leetCode.model.TreeNode; import cn.whaifree.leetCode.model.TreeNode;
import com.sun.org.apache.regexp.internal.RE;
import org.junit.Test; import org.junit.Test;
import java.util.Deque; import java.util.Deque;

14
src/main/java/cn/whaifree/leetCode/Tree/LeetCode513.java
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@ -1,12 +1,14 @@
package cn.whaifree.leetCode.Tree; package cn.whaifree.leetCode.Tree;
import cn.whaifree.leetCode.model.TreeNode; import cn.whaifree.leetCode.model.TreeNode;
import javafx.scene.layout.VBox;
import org.junit.Test; import org.junit.Test;
import java.util.ArrayList; import java.util.ArrayList;
import java.util.Deque; import java.util.Deque;
import java.util.List; import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
/** /**
* @version 1.0 * @version 1.0
@ -17,6 +19,16 @@ import java.util.List;
public class LeetCode513 { public class LeetCode513 {
@Test @Test
public void test() { public void test() {
ExecutorService executor = Executors.newFixedThreadPool(5);
executor.submit(new Callable<Object>() {
@Override
public Object call() throws Exception {
return null;
}
});
TreeNode root = TreeNode.constructTree(new Integer[]{1}); TreeNode root = TreeNode.constructTree(new Integer[]{1});
root.printTree(); root.printTree();
System.out.println(new Solution1().findBottomLeftValue(root)); System.out.println(new Solution1().findBottomLeftValue(root));

2
src/main/java/cn/whaifree/leetCode/Tree/LeetCode654.java
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@ -1,7 +1,7 @@
package cn.whaifree.leetCode.Tree; package cn.whaifree.leetCode.Tree;
import cn.whaifree.leetCode.model.TreeNode; import cn.whaifree.leetCode.model.TreeNode;
import jdk.nashorn.internal.runtime.RewriteException;
import org.junit.Test; import org.junit.Test;
/** /**

34
src/main/java/cn/whaifree/redo/redoAll/LeetCode203.java Normal file
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@ -0,0 +1,34 @@
package cn.whaifree.redo.redoAll;
import cn.whaifree.leetCode.model.ListNode;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/20 14:01
* @注释
*/
public class LeetCode203 {
@Test
public void main() {
new Solution().removeElements(ListNode.listNodeFromArray(new int[]{1, 2, 6, 3, 4, 5, 6}), 6).printList();
}
class Solution {
public ListNode removeElements(ListNode head, int val) {
if (head == null) {
return null;
}
head.next = removeElements(head.next, val);
if (head.val == val) {
return head.next;
}
return head;
}
}
}

35
src/main/java/cn/whaifree/redo/redoAll/LeetCode206.java Normal file
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@ -0,0 +1,35 @@
package cn.whaifree.redo.redoAll;
import cn.whaifree.leetCode.model.ListNode;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/20 14:06
* @注释
*/
public class LeetCode206 {
@Test
public void test() {
ListNode listNode = new Solution().reverseList(ListNode.listNodeFromArray(new int[]{1, 2, 3, 4, 5}));
listNode.printList();
}
class Solution {
public ListNode reverseList(ListNode head) {
return reverse(null, head);
}
public ListNode reverse(ListNode before, ListNode after) {
if (after == null) {
return before;
}
ListNode tmp = after.next;
after.next = before;
return reverse(after, tmp);
}
}
}

4
src/main/java/cn/whaifree/redo/redo_24_2_22/LeetCode206.java
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@ -1,12 +1,8 @@
package cn.whaifree.redo.redo_24_2_22; package cn.whaifree.redo.redo_24_2_22;
import cn.whaifree.leetCode.model.ListNode; import cn.whaifree.leetCode.model.ListNode;
import cn.whaifree.leetCode.model.TreeNode;
import com.sun.xml.internal.ws.server.sei.ValueGetter;
import org.junit.Test; import org.junit.Test;
import java.util.List;
/** /**
* @version 1.0 * @version 1.0
* @Author whai文海 * @Author whai文海

1
src/main/java/cn/whaifree/redo/redo_24_3_1/LeetCode45.java
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@ -2,7 +2,6 @@ package cn.whaifree.redo.redo_24_3_1;
import org.junit.Test; import org.junit.Test;
import javax.annotation.PreDestroy;
/** /**
* @version 1.0 * @version 1.0

44
src/main/java/cn/whaifree/redo/redo_24_3_16/LeetCode236.java Normal file
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@ -0,0 +1,44 @@
package cn.whaifree.redo.redo_24_3_16;
import cn.whaifree.leetCode.model.TreeNode;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/17 15:56
* @注释
*/
public class LeetCode236 {
@Test
public void test() {
new Solution().lowestCommonAncestor(TreeNode.constructTreeByArray(1, 2, 3, 4, 5, 6), new TreeNode(4), new TreeNode(5)).printTree();
}
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
if (root.val == p.val || root.val == q.val) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) {
return root;
}
if (left == null) {
return right;
}else {
return left;
}
}
}
}

44
src/main/java/cn/whaifree/redo/redo_24_3_16/LeetCode376.java Normal file
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package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/17 16:09
* @注释
*/
public class LeetCode376 {
@Test
public void test() {
int[] nums = {0,0,0};
int res = new Solution().wiggleMaxLength(nums);
System.out.println(res);
}
class Solution {
public int wiggleMaxLength(int[] nums) {
if (nums.length < 2) {
return 1;
}
// 记录前一个的正负
// 后一个如果和前一个是相反的 res++ 并且相反值变换
// 不是相反的 不处理
int pre = 0;
int cur = 0;
int res = 1;
for (int i = 1; i < nums.length; i++) {
cur = nums[i] - nums[i - 1];
// pre=0 确保第一次能够进去
if ((cur > 0 && pre <= 0) || (cur < 0 && pre >= 0)) {
res++;
pre = cur;
}
}
return res;
}
}
}

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src/main/java/cn/whaifree/redo/redo_24_3_16/LeetCode416.java Normal file
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package cn.whaifree.redo.redo_24_3_16;
import org.junit.Test;
/**
* @version 1.0
* @Author whai文海
* @Date 2024/3/17 17:04
* @注释
*/
public class LeetCode416 {
@Test
public void test() {
int[] nums = {1, 5, 11, 5};
boolean canPartition = new Solution().canPartition(nums);
System.out.println(canPartition);
}
class Solution {
/**
* 0-1背包问题
*
* dp[i][j] 表示 放入物品i后，容量为j的背包的最大价值
*
* 背包容量为sum/2
* 物品重量为nums[i]
* 物品价值为nums[i]
*
* [3,5,11,5]
*
* 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1
* 0 0 0 0 3 3
* 1
* 2
*
*
*
* @param nums
* @return
*/
public boolean canPartition(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
if (sum % 2 != 0) {
return false;
}
int half = sum / 2;
int length = nums.length;
int[][] dp = new int[length][half+1];
for (int i = nums[0]; i <= half; i++) {
dp[0][i] = nums[0];
}
// dp[i][j] 表示 放入物品i后，容量为j的背包的最大价值
for (int i = 1; i < length; i++) {
for (int j = 1; j <= half; j++) {
// 放的下
if (j >= nums[i]) {
dp[i][j] = Math.max(
dp[i - 1][j],
dp[i - 1][j - nums[i]] + nums[i]
);
}else {
dp[i][j] = dp[i - 1][j];
}
// 放不下
if (dp[i][j] == half) {
return true;
}
}
}
return false;
}
}
class Solution1 {
/**
* 将nums的元素放入背包中
* 1. 背包容量 sum/2
* 2. 物品i重量nums[i]；价值nums[i]
*
* 0 1 2 3 4 5
* 0
* 1
* 2
*
* dp[j]表示 j背包容量的最大价值
*
*
*
* // 递推公式
* 2. 放下 dp[j] = max（dp[j], dp[j-weight[i]]+nums[i]）
*
* 从后往前循环，每次取得状态不会和之前取得状态重合，这样每种物品就只取一次了。
*
* @param nums
* @return
*/
public boolean canPartition(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
if (sum % 2 != 0) {
return false;
}
// 包裹最大容量
int packageCapacity = sum / 2;
int[] dp = new int[packageCapacity + 1];
for (int i = 0; i < nums.length; i++) {
// 包的容量必须大于物品i重量才能放进去
/**
* 如果正序遍历
* dp[1] = dp[1 - weight[0]] + value[0] = 15
* 此时容量1的背包已经放入了
* dp[2] = dp[2 - weight[0]] + value[0] = 30
* 此时又放入了一次
*/
for (int j = packageCapacity; j >= nums[i]; j--) {
dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
//剪枝一下，每一次完成內層的for-loop，立即檢查是否dp[target] == target，優化時間複雜度（26ms -> 20ms）
if(dp[packageCapacity] == packageCapacity)
return true;
}
}
// 包裹容量和目标值一样
return dp[packageCapacity] == packageCapacity;
}
}
}