diff --git a/src/main/java/cn/whaifree/leetCode/Array/LeetCode31.java b/src/main/java/cn/whaifree/leetCode/Array/LeetCode31.java
new file mode 100644
index 0000000..00147ad
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/Array/LeetCode31.java
@@ -0,0 +1,161 @@
+package cn.whaifree.leetCode.Array;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/5 13:18
+ * @注释
+ */
+public class LeetCode31 {
+ @Test
+ public void test() {
+ int[] nums = {3,2,1};
+ new Solution1().nextPermutation(nums);
+
+ for (int i = 0; i < nums.length; i++) {
+ System.out.print(nums[i] + " ");
+ }
+
+ }
+
+ class Solution {
+ public void nextPermutation(int[] nums) {
+ if (nums.length == 1) {
+ return;
+ }
+
+ // 1.下一个数 比当前数大
+ // 将后面的「大数」与前面的「小数」交换，就能得到更大的数
+ // 2. 下一个数 增加的幅度尽可能的小
+ // - 尽可能靠右的低位 进行交换
+ // - 将一个 尽可能小的「大数」 与前面的「小数」交换
+ // 123465 把 5和4换
+ // - 「大数」后面的所有数 重置为升序
+ // 123564 把5后面重新排序
+
+ // 1. 从后往前找到第一个升序排列，此时后面那部分一定是降序
+ int i;
+ for (i = nums.length - 2; i >= 0; i--) {
+ if (nums[i] < nums[i + 1]) {
+ break;
+ }
+ }
+ if (i == -1) {
+ // 最后一个排序 654321，直接逆转
+ reverse(nums, 0, nums.length - 1);
+ return;
+ }
+
+ // 2. 从i+1开始找到最小的值
+ int min;
+ for (min = nums.length - 1; min > 0; min--) {
+ if (nums[min] > nums[i]) {
+ break;
+ }
+ }
+ // 3. 交换i和minIndex
+ swap(nums, i, min);
+ // 4. 后面为降序，直接让其逆转变为升序
+ reverse(nums, i + 1, nums.length - 1);
+
+
+ }
+
+ public void reverse(int[] nums, int start, int end) {
+ while (start < end) {
+ swap(nums, start, end);
+ start++;
+ end--;
+ }
+ }
+
+ public void swap(int[] nums, int i, int j) {
+ int temp = nums[i];
+ nums[i] = nums[j];
+ nums[j] = temp;
+ }
+ }
+
+ class Solution1 {
+ public void nextPermutation(int[] nums) {
+ if (nums.length == 1) {
+ return;
+ }
+
+ int i = nums.length - 2;
+ while (i >= 0) {
+ if (nums[i] < nums[i + 1]) {
+ break;
+ }
+ i--;
+ }
+
+ if (i == -1) {
+ // 最后一个排序 654321，直接逆转
+ reverse(nums, 0, nums.length - 1);
+ return;
+ }
+
+ int min = nums.length - 1;
+ while (min > 0) {
+ if (nums[min] > nums[i]) {
+ break;
+ }
+ min--;
+ }
+
+
+ // 3. 交换i和minIndex
+ swap(nums, i, min);
+ // 4. 后面为降序，直接让其逆转变为升序
+ reverse(nums, i + 1, nums.length - 1);
+ }
+
+ public void reverse(int[] nums, int start, int end) {
+ while (start < end) {
+ swap(nums, start, end);
+ start++;
+ end--;
+ }
+ }
+
+ public void swap(int[] nums, int i, int j) {
+ int temp = nums[i];
+ nums[i] = nums[j];
+ nums[j] = temp;
+ }
+ }
+
+
+
+}
+
+class ass extends as{
+
+ @Override
+ public void test() {
+
+ }
+
+ @Override
+ void test1() {
+
+ }
+
+ @Override
+ protected void test2() {
+
+ }
+}
+
+
+abstract class as{
+
+ abstract public void test() ;
+
+ abstract void test1();
+
+ abstract protected void test2();
+}
diff --git a/src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode32.java b/src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode32.java
new file mode 100644
index 0000000..b3ea311
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/LeetCode/LeetCode32.java
@@ -0,0 +1,155 @@
+package cn.whaifree.leetCode.LeetCode;
+
+import org.junit.Test;
+
+import java.util.Deque;
+import java.util.LinkedList;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 11:27
+ * @注释
+ */
+public class LeetCode32 {
+ @Test
+ public void test() {
+ int i = new Solution3().longestValidParentheses("(()");
+ System.out.println(i);
+ }
+
+ class Solution {
+ /**
+ * 找出最长有效（格式正确且连续）括号子串的长度。
+ *
+ * dp[i] 表示从0-i内最长有效括号子串的长度
+ *
+ * if char[i]='(' || retailLeft<0
+ * = dp[i-1]
+ * if char[i]=')'&&retailLeft>0
+ * = dp[i-1]+2
+ *
+ * @param s
+ * @return
+ */
+ public int longestValidParentheses(String s) {
+ char[] chars = s.toCharArray();
+ int[] dp = new int[s.length() + 1];
+ int retailLeft = 0;
+ for (int i = 0; i < chars.length; i++) {
+ if (chars[i] == '(' ) {
+ retailLeft++;
+ dp[i] = dp[i - 1];
+ } else if (chars[i] == ')' && retailLeft > 0) {
+
+ }
+ }
+
+
+ return 0;
+ }
+
+ }
+
+ class Solution1 {
+ /**
+ * ()(()
+ *
+ * ...
+ *
+ * 1.匹配成功 用栈匹配，把所有匹配成功的flag设置为1
+ * 2.最长连续 统计flag中最长连续1的长度
+ *
+ * @param s
+ * @return
+ */
+ public int longestValidParentheses(String s) {
+ int[] flag = new int[s.length()];
+ Deque stack = new LinkedList<>();
+
+ for (int i = 0; i < s.length(); i++) {
+ if (s.charAt(i) == '(') {
+ stack.push(i);
+ } else if (!stack.isEmpty()) {
+ Integer pop = stack.pop();
+ if (s.charAt(pop) == '(') {
+ flag[i] = 1;
+ flag[pop] = 1;
+ }
+ }
+ }
+
+ // 计算flag中最长连续出现1的次数
+ int len = 0;
+ int maxLen = Integer.MIN_VALUE;
+ for (int i = 0; i < flag.length; i++) {
+ if (flag[i] == 1) {
+ len += 1;
+ }else {
+ maxLen = Math.max(maxLen,len);
+ len = 0;
+ }
+ }
+
+ return Math.max(maxLen, len); // (() 这个用例
+ }
+ }
+ class Solution3 {
+ /**
+ * 使用两个计数器
+ * - 当两个计数器相等 则检查最长匹配
+ * - 当右边比左边还多，重置
+ *
+ * 需要 从前往后+从后往前
+ *
+ * @param s
+ * @return
+ */
+ public int longestValidParentheses(String s) {
+ int left = 0, right = 0, maxlength = 0;
+ // 左右计数器，
+ // - 一旦右边计数比左边大的时候，重置
+ // - 左右计数器相等的时候，匹配maxLength
+
+ for (int i = 0; i < s.length(); i++) {
+ if (s.charAt(i) == '(') {
+ left++;
+ } else if (s.charAt(i) == ')') {
+ right++;
+ }
+
+ if (left == right) {
+ maxlength = Math.max(maxlength, left + right);
+ } else if (right > left) {
+ // 右边比左边还多，重置，重新计算
+ right = 0;
+ left = 0;
+ }
+ }
+ // 重新从下一个字符开始计算，但这样会漏掉一种情况，就是遍历的时候左括号的数量始终大于右括号的数量，即 (() ，这种时候最长有效括号是求不出来的。
+
+ // 只要从左到右再来一次，
+
+ left = 0;
+ right = 0;
+ for (int i = s.length() - 1; i > 0; i--) {
+ if (s.charAt(i) == '(') {
+ left++;
+ } else if (s.charAt(i) == ')') {
+ right++;
+ }
+
+ if (left == right) {
+ maxlength = Math.max(maxlength, left + right);
+ } else if (left > right) {
+ // 右边比左边还多，重置，重新计算
+ right = 0;
+ left = 0;
+ }
+ }
+
+ return maxlength;
+ }
+ }
+
+}
diff --git a/src/main/java/cn/whaifree/leetCode/LinkedList/LeetCode23.java b/src/main/java/cn/whaifree/leetCode/LinkedList/LeetCode23.java
new file mode 100644
index 0000000..910aa3c
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/LinkedList/LeetCode23.java
@@ -0,0 +1,129 @@
+package cn.whaifree.leetCode.LinkedList;
+
+import cn.whaifree.leetCode.model.ListNode;
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/5 12:00
+ * @注释
+ */
+public class LeetCode23 {
+
+ @Test
+ public void test()
+ {
+ new Solution2().mergeKLists(
+ new ListNode[]{
+ ListNode.listNodeFromArray(new int[]{}),
+ ListNode.listNodeFromArray(new int[]{1, 3, 4}),
+ ListNode.listNodeFromArray(new int[]{2, 6})
+ }
+ ).printList();
+ }
+
+ class Solution {
+ public ListNode mergeKLists(ListNode[] lists) {
+ if (lists.length == 0) {
+ return null;
+ }
+ List arrayList = new ArrayList<>(Arrays.asList(lists));
+ return merge(arrayList);
+ }
+
+ public ListNode merge(List lists) {
+ // 删除list中所有null
+ // 使用removeIf()方法删除所有null元素
+ lists.removeIf(Objects::isNull);
+
+ if (lists.size() == 0) {
+ return null;
+ }
+
+ int minIndex = 0;
+ // 找到最小的返回
+ // 这个最小的的下一个就是递归的
+
+ // 找到最小的头节点的位置
+ for (int i = 0; i < lists.size(); i++) {
+ if (lists.get(minIndex).val > lists.get(i).val) {
+ minIndex = i;
+ }
+ }
+ ListNode minNode = lists.get(minIndex);
+ // 去掉本节点
+ lists.set(minIndex, minNode.next);
+ minNode.next = merge(lists);
+ return minNode;
+
+ }
+
+ }
+
+
+ class Solution1 {
+ public ListNode mergeKLists(ListNode[] lists) {
+ for (int i = 1; i < lists.length; i++) {
+ lists[i] = merge(lists[i], lists[i - 1]);
+ }
+ return lists[lists.length - 1];
+ }
+
+ public ListNode merge(ListNode one, ListNode two) {
+ if (one == null) {
+ return two;
+ } else if (two == null){
+ return one;
+ }
+
+ if (one.val < two.val) {
+ one.next = merge(one.next, two);
+ return one;
+ }else {
+ two.next = merge(one, two.next);
+ return two;
+ }
+ }
+
+ }
+
+ class Solution2 {
+ // 时间复杂度：
+ // O(nlogk)，其中 k 为 lists的长度，n 为所有链表的节点数之和。
+
+ //著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
+ public ListNode mergeKLists(ListNode[] lists) {
+
+
+ PriorityQueue queue = new PriorityQueue<>(new Comparator() {
+ @Override
+ public int compare(ListNode o1, ListNode o2) {
+ return o1.val - o2.val;
+ }
+ });
+
+ for (ListNode list : lists) {
+ if (list!=null) queue.add(list);
+ }
+
+ ListNode head = new ListNode(-1);
+ ListNode headIndex = head;
+ while (!queue.isEmpty()) {
+ ListNode poll = queue.poll();
+ headIndex.next = poll;
+ headIndex = headIndex.next;
+ if (poll.next != null) {
+ queue.add(poll.next);
+ }
+ }
+
+ return head.next;
+ }
+
+
+ }
+
+}
diff --git a/src/main/java/cn/whaifree/leetCode/String/LeetCode10.java b/src/main/java/cn/whaifree/leetCode/String/LeetCode10.java
new file mode 100644
index 0000000..28d1c84
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/String/LeetCode10.java
@@ -0,0 +1,39 @@
+package cn.whaifree.leetCode.String;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/5 11:36
+ * @注释
+ */
+public class LeetCode10 {
+
+ class Solution {
+ public boolean isMatch(String s, String p) {
+
+ char[] sChar = s.toCharArray();
+ char[] pChar = p.toCharArray();
+ /**
+ * dp[sIndex][pIndex]表示s字符串的前sIndex个字符是否和p字符串的前pIndex个字符完成正则表达式的匹配
+ *
+ * 当p字符串的第pIndex个字符不为*时
+ * if p第pIndex个字符与s的第sIndex个字符匹配
+ * dp[sIndex][pIndex] =
+ * f[sIndex - 1][pIndex - 1](即：s的前sIndex - 1个字符和p的前pIndex-1个字符的匹配结果)。
+ * 如果p第pIndex个字符与s的第sIndex个字符不匹配，
+ * 则结果为false；
+ * 当p字符串的第pIndex个字符为*时
+ * 此时*前面的符号可以出现0次或任意次
+ * - x出现0次
+ * 此时结果为f[sIndex][pIndex - 2]。此时相当于删除了*以及x
+ * - x出现非0的任意次， 此时的结果为f[sIndex - 1][pIndex] && (p.charAt(pIndex - 2) == '.' || p.charAt(pIndex - 2) == s.charAt(sIndex - 1));
+ *
+ */
+
+
+ return false;
+
+ }
+ }
+
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode209.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode209.java
new file mode 100644
index 0000000..1d6f22d
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode209.java
@@ -0,0 +1,73 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 15:07
+ * @注释
+ */
+public class LeetCode209 {
+ @Test
+ public void test() {
+
+ int[] nums = {2, 3, 1, 2, 4, 3};
+ int target = 7;
+ int i = new Solution2().minSubArrayLen(target, nums);
+ System.out.println(i);
+ }
+
+ class Solution {
+ public int minSubArrayLen(int target, int[] nums) {
+ int right = 0;
+ int left = 0;
+ int sum = 0;
+ int minLength = Integer.MAX_VALUE;
+ while (right < nums.length) {
+ sum += nums[right];
+ while (left <= right && sum >= target) {
+ minLength = Math.min(minLength, right - left + 1);
+ sum -= nums[left];
+ left++;
+ }
+ right++;
+ }
+ return minLength == Integer.MAX_VALUE ? 0 : minLength;
+ }
+ }
+
+ class Solution2 {
+ /**
+ * 
+ * @param target
+ * @param nums
+ * @return
+ */
+ public int minSubArrayLen(int target, int[] nums) {
+
+ // 前缀和
+ int[] preSum = new int[nums.length + 1];
+ for (int i = 1; i < preSum.length; i++) {
+ preSum[i] = preSum[i - 1] + nums[i - 1];
+ }
+
+
+ int minLen = Integer.MAX_VALUE;
+ for (int i = 0; i < preSum.length; i++) {
+ // preSum[right] = target + preSum[i]
+ int fill = target + preSum[i];
+ int right = Arrays.binarySearch(preSum, fill);
+ if (right < 0) {
+ right = -right - 1;
+ }
+ if (right < preSum.length) {
+ minLen = Math.min(minLen, right - i);
+ }
+ }
+ return minLen == Integer.MAX_VALUE ? 0 : minLen;
+ }
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode279.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode279.java
new file mode 100644
index 0000000..290f320
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode279.java
@@ -0,0 +1,78 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 16:07
+ * @注释
+ */
+public class LeetCode279 {
+
+ @Test
+ public void test()
+ {
+ Solution solution = new Solution();
+ int i = solution.numSquares3(1);
+ System.out.println(i);
+ }
+
+ class Solution {
+ public int numSquares(int n) {
+
+ // dp[j] 表示从[0-i*i]和为j的完全平方数的数量
+ // dp[j] = dp[j-i*i]+1
+ int[] dp = new int[n+1];
+ int sqrt = (int) Math.sqrt(n);
+ Arrays.fill(dp, Integer.MAX_VALUE);
+
+
+ dp[0] = 0;
+ for (int j = 1; j < n + 1; j++) {
+ for (int i = 1; i <= sqrt; i++) {
+ if (j >= i * i) {
+ dp[j] = Math.min(dp[j], dp[j - i * i] + 1);
+ }
+ }
+ }
+ return dp[n];
+ }
+
+ public int numSquares1(int n) {
+
+ // dp[j] 表示从[0-i*i]和为j的完全平方数的数量
+ // dp[j] = dp[j-i*i]+1
+ int[] dp = new int[n+1];
+ Arrays.fill(dp, Integer.MAX_VALUE);
+
+
+ dp[0] = 0;
+ for (int j = 1; j < n + 1; j++) {
+ for (int i = 1; i * i <= j; i++) {
+ dp[j] = Math.min(dp[j], dp[j - i * i] + 1);
+ }
+ }
+ return dp[n];
+ }
+
+ public int numSquares3(int n) {
+
+ // dp[j] 表示从[0-i*i]和为j的完全平方数的数量
+ // dp[j] = dp[j-i*i]+1
+ int[] dp = new int[n+1];
+ Arrays.fill(dp, Integer.MAX_VALUE);
+
+
+ dp[0] = 0;
+ for (int i = 1; i <= n; i++) {
+ for (int j = i * i; j <= n; j++) {
+ dp[j] = Math.min(dp[j], dp[j - i * i] + 1);
+ }
+ }
+ return dp[n];
+ }
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode322.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode322.java
new file mode 100644
index 0000000..50bb77c
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode322.java
@@ -0,0 +1,49 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 16:28
+ * @注释
+ */
+public class LeetCode322 {
+ @Test
+ public void test() {
+ int[] coins = { 2 };
+ int amount = 3;
+ Solution solution = new Solution();
+ int i = solution.coinChange(coins, amount);
+ System.out.println(i);
+ }
+
+ class Solution {
+ /**
+ * 背包容量amount+1
+ *
+ * dp[j] 表示从0-i中任意取放入amount
+ * dp[j] = math.min(dp[j] , dp[j-coins[i]]+1)
+ *
+ *
+ * @param coins
+ * @param amount
+ * @return
+ */
+ public int coinChange(int[] coins, int amount) {
+ int[] dp = new int[amount + 1];
+ Arrays.fill(dp, Integer.MAX_VALUE);
+ dp[0] = 0;
+ for (int i = 0; i < coins.length; i++) {
+ for (int j = coins[i]; j < amount + 1; j++) {
+ if (dp[j - coins[i]] != Integer.MAX_VALUE) {
+ dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1);
+ }
+ }
+ }
+ return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
+ }
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode377.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode377.java
new file mode 100644
index 0000000..b1240f7
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode377.java
@@ -0,0 +1,45 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 17:00
+ * @注释
+ */
+public class LeetCode377 {
+
+ @Test
+ public void test() {
+ Solution solution = new Solution();
+ int[] nums = {9};
+ int target = 3;
+ int i = solution.combinationSum4(nums, target);
+ System.out.println(i);
+ }
+
+ class Solution {
+ public int combinationSum4(int[] nums, int target) {
+ int[] dp = new int[target + 1];
+
+ dp[0] = 1;
+
+ for (int j = 0; j < target + 1; j++) {
+ for (int i = 0; i < nums.length; i++) {
+ if (nums[i] <= j) {
+ dp[j] = dp[j] + dp[j - nums[i]];
+ }
+ }
+ }
+
+// for (int i = 0; i < nums.length; i++) {
+// for (int j = nums[i]; j <= target; j++) {
+// dp[j] = dp[j] + dp[j - nums[i]];
+// }
+// }
+
+ return dp[target];
+ }
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode474.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode474.java
new file mode 100644
index 0000000..c576bc7
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode474.java
@@ -0,0 +1,74 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 17:08
+ * @注释
+ */
+public class LeetCode474 {
+
+ @Test
+ public void test() {
+ String[] strs = new String[]{"10", "0001", "111001", "1", "0"};
+ int m = 5;
+ int n = 3;
+ System.out.println(new Solution().findMaxForm(strs, m, n));
+ }
+
+ class Solution {
+ public int findMaxForm(String[] strs, int m, int n) {
+
+ /**
+ * dp[i][j][k] 表示包含 i个0，j个1，从前k个子集中取出的最大子集长度
+ *
+ * i j 为背包容量
+ * k为物品
+ *
+ * for 背包
+ * 统计0，1数量
+ * fori
+ * forj
+ * if(i>=zero j>=one)
+ * dp[i][j][k] = max(dp[i][j][k-1],dp[i-zero][j-one][k-1]+1)
+ * else
+ * dp[i][j][k] = dp[i][j][k-1]
+ *
+ */
+
+
+ int[][][] dp = new int[m + 1][n + 1][strs.length + 1];
+
+ for (int k = 1; k <= strs.length; k++) {
+ String str = strs[k-1];
+ int zero = calculateZeroNumber(str);
+ int one = str.length() - zero;
+ for (int i = 0; i <= m; i++) {
+ for (int j = 0; j <= n; j++) {
+ if (i >= zero && j >= one) {
+ dp[i][j][k] = Math.max(dp[i][j][k - 1], dp[i - zero][j - one][k - 1] + 1);
+ } else {
+ dp[i][j][k] = dp[i][j][k - 1];
+ }
+ }
+ }
+ }
+
+ return dp[m][n][strs.length];
+ }
+
+ public int calculateZeroNumber(String str) {
+ int num = 0;
+ for (int i = 0; i < str.length(); i++) {
+ if (str.charAt(i) == '0') {
+ num++;
+ }
+ }
+ return num;
+ }
+
+
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode494.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode494.java
new file mode 100644
index 0000000..8b569e1
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode494.java
@@ -0,0 +1,47 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 17:34
+ * @注释
+ */
+public class LeetCode494 {
+
+ @Test
+ public void test()
+ {
+ Solution solution = new Solution();
+ int[] nums = new int[]{1, 1, 1, 1, 1};
+ int target = 3;
+ int i = solution.findTargetSumWays(nums, target);
+ System.out.println(i);
+ }
+
+ class Solution {
+ public int findTargetSumWays(int[] nums, int target) {
+ int sum = 0;
+ for (int i = 0; i < nums.length; i++) {
+ sum += nums[i];
+ }
+ if (Math.abs(target) > sum) return 0; // 此时没有方案
+ if ((target + sum) % 2 == 1) return 0; // 此时没有方案
+
+ int left = (sum + target) / 2;
+
+ // dp[j] 表示 从0-i 中任意取，背包容量为 j 时，能装下的组合数
+ // dp[j] = dp[j] + dp[j-nums[i]]
+ int[] dp = new int[left + 1];
+ dp[0] = 1;
+ for (int i = 0; i < nums.length; i++) {
+ for (int j = left; j >= nums[i]; j--) {
+ // 已经有一个1（nums[i]） 的话，有 dp[4]种方法 凑成 容量为5的背包。
+ dp[j] = dp[j] + dp[j - nums[i]];
+ }
+ }
+ return dp[left];
+ }
+ }
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode53.java b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode53.java
new file mode 100644
index 0000000..19fe657
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_1/LeetCode53.java
@@ -0,0 +1,126 @@
+package cn.whaifree.redo.redo_24_4_1;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 12:44
+ * @注释
+ */
+public class LeetCode53 {
+
+ @Test
+ public void test() {
+
+ int[] nums = new int[]{1};
+ int maxSubArray = new Solution1().maxSubArray(nums);
+ System.out.println(maxSubArray);
+ }
+
+ class Solution {
+ public int maxSubArray(int[] nums) {
+ // 滑动窗口
+ // 遇到和为负数就一直缩小left
+
+ int maxSum = Integer.MIN_VALUE;
+ int nowSum = 0;
+ for (int i = 0; i < nums.length; i++) {
+ nowSum += nums[i];
+ maxSum = Math.max(maxSum, nowSum);
+ if (nowSum < 0) {
+ nowSum = 0;
+ }
+ }
+ return maxSum;
+ }
+ }
+
+
+ class Solution1 {
+ /**
+ * dp
+ *
+ * dp[i] 表示i处结尾的的最大连续子数组的和
+ *
+ * dp[0] = nums[0]
+ * dp[i]
+ * if nums[i]+dp[i-1] > 0
+ * dp[i] = dp[i-1]+nums[i]
+ * else
+ * dp[i] = nums[i]
+ *
+ * @param nums
+ * @return
+ */
+ public int maxSubArray(int[] nums) {
+ int[] dp = new int[nums.length];
+ dp[0] = nums[0];
+ int max = nums[0];
+ for (int i = 1; i < nums.length; i++) {
+ if (dp[i - 1] > 0 && nums[i] + dp[i - 1] > 0) {
+ dp[i] = dp[i - 1] + nums[i];
+ } else {
+ dp[i] = nums[i];
+ }
+ max = Math.max(max, dp[i]);
+ }
+ return max;
+ }
+ }
+
+ class Solution4 {
+ /**
+ * 前缀和
+ * @param nums
+ * @return
+ */
+// public int maxSubArray(int[] nums) {
+//
+// int[] preSum = new int[nums.length + 1];
+// for (int i = 1; i < nums.length; i++) {
+// preSum[i] = preSum[i - 1] + nums[i - 1];
+// }
+//
+//
+// }
+ }
+
+
+}
+
+class XieChen {
+
+ /**
+ * 将偶数的区间/2 让整个区间最大
+ * 给定一个数组，对一个区间（区间内全部都是偶数）全部/2，使得整个数组和最大，求数组最大和。
+ */
+ public static void main(String[] args) {
+
+
+
+ }
+
+ public void xc(int[] nums) {
+
+ // 前缀和
+ int[] preSum = new int[nums.length + 1];
+ for (int i = 1; i < nums.length; i++) {
+ preSum[i] = preSum[i - 1] + nums[i - 1];
+ }
+
+ int minLoc = 0;
+
+ int left = 0;
+ int right = 0;
+ while (right < nums.length) {
+
+
+ right++;
+ }
+
+
+ }
+
+
+}
diff --git a/src/main/java/cn/whaifree/redo/redo_24_4_6/LeetCode11.java b/src/main/java/cn/whaifree/redo/redo_24_4_6/LeetCode11.java
new file mode 100644
index 0000000..21fa75a
--- /dev/null
+++ b/src/main/java/cn/whaifree/redo/redo_24_4_6/LeetCode11.java
@@ -0,0 +1,42 @@
+package cn.whaifree.redo.redo_24_4_6;
+
+import org.junit.Test;
+
+/**
+ * @version 1.0
+ * @Author whai文海
+ * @Date 2024/4/6 12:35
+ * @注释
+ */
+public class LeetCode11 {
+
+ @Test
+ public void test() {
+ Solution solution = new Solution();
+ int[] height = {1,1};
+ int i = solution.maxArea(height);
+ System.out.println(i);
+ }
+
+ class Solution {
+ public int maxArea(int[] height) {
+
+ // capacity = (right-left)*min(right,left)
+ // 移动短的，可能变大，不变，变小
+ // 移动长的，只会变小
+ int left = 0;
+ int right = height.length - 1;
+ int capacity = Integer.MIN_VALUE;
+ while (left < right) {
+ capacity = Math.max(capacity, (right - left) * Math.min(height[left], height[right]));
+ if (height[left] < height[right]) {
+ left++;
+ }else {
+ right--;
+ }
+ }
+ return capacity;
+ }
+ }
+
+}