- 239 队列的应用，没做出来
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whai
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src/main/java/cn/whaifree/leetCode/Stack/LeetCode1047.java
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64
src/main/java/cn/whaifree/leetCode/Stack/LeetCode1047.java
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package cn.whaifree.leetCode.Stack;
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import org.junit.Test;
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import java.util.ArrayDeque;
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import java.util.Deque;
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import java.util.LinkedList;
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import java.util.function.Consumer;
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public class LeetCode1047 {
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@Test
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public void test() {
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System.out.println(new Solution1().removeDuplicates("adff"));
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}
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class Solution {
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public String removeDuplicates(String s) {
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Deque<Character> stack = new ArrayDeque<>();
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char[] chars = s.toCharArray();
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for (char aChar : chars) {
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// 入栈前判断第一个是不是相同的
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if (!stack.isEmpty() && stack.peek() == aChar) {
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stack.pop();
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}else {
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stack.push(aChar);
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}
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}
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char[] ans = new char[stack.size()];
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for (int i = stack.size() - 1; i >=0; i--) {
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ans[i] = stack.pop();
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}
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return new String(ans);
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}
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}
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class Solution1 {
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/**
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* stringBuilder也能作为栈使用
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*/
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public String removeDuplicates(String s) {
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// 将StringBuilder作为栈 abbacd
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StringBuilder res = new StringBuilder();
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int top = -1;
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for (int i = 0; i < s.length(); i++) {
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char c = s.charAt(i);
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if (res.length() != 0 && c == res.charAt(top)) {
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res.deleteCharAt(top);
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top--;
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} else {
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res.append(c);
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top++;
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}
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}
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return res.toString();
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}
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}
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}
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90
src/main/java/cn/whaifree/leetCode/Stack/LeetCode150.java
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90
src/main/java/cn/whaifree/leetCode/Stack/LeetCode150.java
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package cn.whaifree.leetCode.Stack;
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import org.junit.Test;
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import java.util.Deque;
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import java.util.LinkedList;
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/**
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* 逆转波兰表达式
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*/
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public class LeetCode150 {
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@Test
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public void test() {
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System.out.println(new Solution1().evalRPN(new String[]{"2", "1", "+", "3", "*"}));
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System.out.println(new Solution1().evalRPN(new String[]{"4", "13", "5", "/", "+"}));
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}
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class Solution {
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public int evalRPN(String[] tokens) {
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Deque<String> stack = new LinkedList<>();
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int index = 0;
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while (index < tokens.length) {
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String token = tokens[index];
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if (token.equals("+")) {
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Integer pop1 = Integer.parseInt(stack.pop());
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Integer pop2 = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(pop1 + pop2));
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} else if (token.equals("-")) { //避免多次判断
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Integer pop1 = Integer.parseInt(stack.pop());
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Integer pop2 = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(pop2 - pop1));
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} else if (token.equals("*")) {
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Integer pop1 = Integer.parseInt(stack.pop());
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Integer pop2 = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(pop2 * pop1));
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} else if (token.equals("/")) {
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Integer pop1 = Integer.parseInt(stack.pop());
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Integer pop2 = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(pop2 / pop1));
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} else {
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stack.push(token);
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}
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index++;
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}
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return Integer.parseInt(stack.pop());
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}
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}
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class Solution1 {
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public int evalRPN(String[] tokens) {
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Deque<String> stack = new LinkedList<>();
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for (String token : tokens) {
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switch (token) {
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case "+":
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int i = Integer.parseInt(stack.pop());
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int j = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(i + j));
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break;
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case "-":
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int k = Integer.parseInt(stack.pop());
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int l = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(l - k));
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break;
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case "*":
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int m = Integer.parseInt(stack.pop());
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int n = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(n * m));
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break;
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case "/":
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int o = Integer.parseInt(stack.pop());
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int p = Integer.parseInt(stack.pop());
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stack.push(String.valueOf(p / o));
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break;
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default:
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stack.push(token);
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break;
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}
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}
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return Integer.parseInt(stack.pop());
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}
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}
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}
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94
src/main/java/cn/whaifree/leetCode/Stack/LeetCode239.java
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94
src/main/java/cn/whaifree/leetCode/Stack/LeetCode239.java
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package cn.whaifree.leetCode.Stack;
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import org.junit.Test;
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import java.util.Deque;
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import java.util.LinkedList;
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/**
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* @Author whai文海
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* @Date 2024/1/9 14:49
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* @注释
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*
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* 239. 滑动窗口最大值
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* 给你一个整数数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。
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*
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* 返回 滑动窗口中的最大值 。
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*
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*
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*
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* 示例 1：
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*
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* 输入：nums = [1,3,-1,-3,5,3,6,7], k = 3
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* 输出：[3,3,5,5,6,7]
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* 解释：
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* 滑动窗口的位置 最大值
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* --------------- -----
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* [1 3 -1] -3 5 3 6 7 3
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* 1 [3 -1 -3] 5 3 6 7 3
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* 1 3 [-1 -3 5] 3 6 7 5
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* 1 3 -1 [-3 5 3] 6 7 5
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* 1 3 -1 -3 [5 3 6] 7 6
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* 1 3 -1 -3 5 [3 6 7] 7
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* 示例 2：
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*
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* 输入：nums = [1], k = 1
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* 输出：[1]
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*/
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public class LeetCode239 {
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@Test
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public void test() {
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int[] nums = new int[]{1, 3, -1, -3, 5, 3, 6, 7};
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int[] res = new Solution().maxSlidingWindow(nums, 3);
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for (int re : res) {
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System.out.println(re);
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}
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}
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class Solution {
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/**
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* 维持一个单调队列，队列里存放的是对应值得下标。
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* 如果 此次队列头部指针为i，即num[i]为目前窗口的最大值。
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* - 如果这个最大值的下标不在下一个窗口的范围内，那么需要将这个值删除
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* @param nums
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* @param k
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* @return
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*/
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public int[] maxSlidingWindow(int[] nums, int k) {
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// 单调队列，维持一个单调递减的队列
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Deque<Integer> queue = new LinkedList<>();
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// n个元素，k个窗口，一共为n-k个输出
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int[] res = new int[nums.length - k + 1];
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for (int i = 0; i < nums.length; i++) {
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// 判断当前队首的值是否在本次的窗口内，i-k为窗口左边
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if (!queue.isEmpty()&&queue.peek() <= i - k) {
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queue.pop();
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}
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// 保证单调递减
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while (!queue.isEmpty() && nums[i] >= nums[queue.peekLast()]) {
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queue.removeLast();
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}
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queue.addLast(i);
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// 从第k个开始才有结果
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if(i+1 >= k){
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res[i+1-k] = nums[queue.peek()];
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}
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// 从第k个开始，先把值存进去，在下次循环判断是否这个值存在于这个新的区间中
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}
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return res;
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}
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}
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}
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