Update

problems/0001.两数之和.md

@@ -352,5 +352,86 @@ List<int> twoSum(List<int> nums, int target) {
 }
 ```
 
+C:
+```c
+
+
+/**
+ * Note: The returned array must be malloced, assume caller calls free().
+ */
+
+// leetcode 支持 ut_hash 函式庫
+
+ typedef struct {
+     int key;
+     int value;
+     UT_hash_handle hh; // make this structure hashable
+ } map;
+
+map* hashMap = NULL;
+
+ void hashMapAdd(int key, int value){
+     map* s;
+     // key already in the hash?
+     HASH_FIND_INT(hashMap, &key, s);
+     if(s == NULL){
+         s = (map*)malloc(sizeof(map));
+         s -> key = key;
+         HASH_ADD_INT(hashMap, key, s);
+     }
+     s -> value = value;
+ }
+
+map* hashMapFind(int key){
+     map* s;
+     // *s: output pointer
+     HASH_FIND_INT(hashMap, &key, s);   
+     return s;
+ }
+
+ void hashMapCleanup(){
+     map* cur, *tmp;
+     HASH_ITER(hh, hashMap, cur, tmp){
+         HASH_DEL(hashMap, cur);
+         free(cur);
+     }
+ }
+
+ void hashPrint(){
+     map* s;
+     for(s = hashMap; s != NULL; s=(map*)(s -> hh.next)){
+         printf("key %d, value %d\n", s -> key, s -> value);
+     }
+ }
+
+ 
+int* twoSum(int* nums, int numsSize, int target, int* returnSize){
+    int i, *ans;
+    // hash find result
+    map* hashMapRes; 
+    hashMap = NULL;
+    ans = malloc(sizeof(int) * 2);
+
+    for(i = 0; i < numsSize; i++){
+        // key 代表 nums[i] 的值，value 代表所在 index;
+        hashMapAdd(nums[i], i);
+    }
+
+    hashPrint();
+
+    for(i = 0; i < numsSize; i++){
+        hashMapRes = hashMapFind(target - nums[i]);
+        if(hashMapRes && hashMapRes -> value != i){
+            ans[0] = i;
+            ans[1] = hashMapRes -> value ;
+            *returnSize = 2;
+            return ans;
+        }
+    }
+    
+    hashMapCleanup();
+    return NULL;
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0019.删除链表的倒数第N个节点.md

@@ -75,7 +75,12 @@ public:
             fast = fast->next;
             slow = slow->next;
         }
-        slow->next = slow->next->next;
+        slow->next = slow->next->next; 
+        
+        // ListNode *tmp = slow->next;  C++释放内存的逻辑
+        // slow->next = tmp->next;
+        // delete nth;
+        
         return dummyHead->next;
     }
 };
@@ -87,30 +92,27 @@ public:
 java:
 
 ```java
-class Solution {
+public ListNode removeNthFromEnd(ListNode head, int n){
-    public ListNode removeNthFromEnd(ListNode head, int n) {
+    ListNode dummyNode = new ListNode(0);
-        ListNode dummy = new ListNode(-1);
+    dummyNode.next = head;
-        dummy.next = head;
 
-        ListNode slow = dummy;
+    ListNode fastIndex = dummyNode;
-        ListNode fast = dummy;
+    ListNode slowIndex = dummyNode;
-        while (n-- > 0) {
-            fast = fast.next;
-        }
-        // 记住 待删除节点slow 的上一节点
-        ListNode prev = null;
-        while (fast != null) {
-            prev = slow;
-            slow = slow.next;
-            fast = fast.next;
-        }
-        // 上一节点的next指针绕过 待删除节点slow 直接指向slow的下一节点
-        prev.next = slow.next;
-        // 释放 待删除节点slow 的next指针, 这句删掉也能AC
-        slow.next = null;
 
-        return dummy.next;
+    //只要快慢指针相差 n 个结点即可
+    for (int i = 0; i < n  ; i++){
+        fastIndex = fastIndex.next;
     }
+
+    while (fastIndex.next != null){
+        fastIndex = fastIndex.next;
+        slowIndex = slowIndex.next;
+    }
+
+    //此时 slowIndex 的位置就是待删除元素的前一个位置。
+    //具体情况可自己画一个链表长度为 3 的图来模拟代码来理解
+    slowIndex.next = slowIndex.next.next;
+    return dummyNode.next;
 }
 ```
 
@@ -341,5 +343,28 @@ object Solution {
   }
 }
 ```
+
+Rust:
+```rust
+impl Solution {
+    pub fn remove_nth_from_end(head: Option<Box<ListNode>>, mut n: i32) -> Option<Box<ListNode>> {
+        let mut dummy_head = Box::new(ListNode::new(0));
+        dummy_head.next = head;
+        let mut fast = &dummy_head.clone();
+        let mut slow = &mut dummy_head;
+        while n > 0 {
+            fast = fast.next.as_ref().unwrap();
+            n -= 1;
+        }
+        while fast.next.is_some() {
+            fast = fast.next.as_ref().unwrap();
+            slow = slow.next.as_mut().unwrap();
+        }
+        slow.next = slow.next.as_mut().unwrap().next.take();
+        dummy_head.next
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0024.两两交换链表中的节点.md

@@ -380,5 +380,51 @@ function swapPairs($head)
 }
 ```
 
+Rust:
+
+```rust
+// 虚拟头节点
+impl Solution {
+    pub fn swap_pairs(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
+        let mut dummy_head = Box::new(ListNode::new(0));
+        dummy_head.next = head;
+        let mut cur = dummy_head.as_mut();
+        while let Some(mut node) = cur.next.take() {
+            if let Some(mut next) = node.next.take() {
+                node.next = next.next.take();
+                next.next = Some(node);
+                cur.next = Some(next);
+                cur = cur.next.as_mut().unwrap().next.as_mut().unwrap();
+            } else {
+                cur.next = Some(node);
+                cur = cur.next.as_mut().unwrap();
+            }
+        }
+        dummy_head.next
+    }
+}
+```
+
+```rust
+// 递归
+impl Solution {
+    pub fn swap_pairs(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
+        if head == None || head.as_ref().unwrap().next == None {
+            return head;
+        }
+
+        let mut node = head.unwrap();
+
+        if let Some(mut next) = node.next.take() {
+            node.next = Solution::swap_pairs(next.next);
+            next.next = Some(node);
+            Some(next)
+        } else {
+            Some(node)
+        }
+    }
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0035.搜索插入位置.md

@@ -429,7 +429,57 @@ function searchInsert($nums, $target)
     return $r + 1;
 }
 ```
-
+### C
+```c
+//版本一 [left, right]左闭右闭区间
+int searchInsert(int* nums, int numsSize, int target){
+    //左闭右开区间 [0 , numsSize-1] 
+        int left =0;
+        int mid =0;
+        int right = numsSize - 1;
+        while(left <= right){//左闭右闭区间 所以可以 left == right
+            mid = left + (right - left) / 2;
+            if(target < nums[mid]){
+        //target 在左区间 [left, mid - 1]中，原区间包含mid，右区间边界可以向左内缩
+                right = mid -1;
+            }else if( target > nums[mid]){
+        //target 在右区间 [mid + 1, right]中,原区间包含mid，左区间边界可以向右内缩
+                left = mid + 1;
+            }else {
+        // nums[mid] == target ，顺利找到target，直接返回mid
+                return mid;
+            }
+        }
+        //数组中未找到target元素
+        //target在数组所有元素之后，[left, right]是右闭区间，需要返回 right +1 
+        return right + 1;
+}
+```
+```c
+//版本二 [left, right]左闭右开区间
+int searchInsert(int* nums, int numsSize, int target){
+    //左闭右开区间 [0 , numsSize) 
+        int left =0;
+        int mid =0;
+        int right = numsSize;
+        while(left < right){//左闭右闭区间 所以 left < right
+            mid = left + (right - left) / 2;
+            if(target < nums[mid]){
+        //target 在左区间 [left, mid)中，原区间没有包含mid，右区间边界不能内缩
+                right = mid ;
+            }else if( target > nums[mid]){
+        // target 在右区间 [mid+1, right)中，原区间包含mid，左区间边界可以向右内缩
+                left = mid + 1;
+            }else {
+        // nums[mid] == target ，顺利找到target，直接返回mid
+                return mid;
+            }
+        }
+	//数组中未找到target元素
+        //target在数组所有元素之后，[left, right)是右开区间， return right即可
+        return right;
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0045.跳跃游戏II.md

@@ -214,7 +214,26 @@ class Solution:
         return ans
 ```
 
+```python
+# 贪心版本二
+class Solution:
+    def jump(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return 0
+        curDistance, nextDistance = 0, 0
+        step = 0
+        for i in range(len(nums)-1):
+            nextDistance = max(nextDistance, nums[i]+i)
+            if i == curDistance:
+                curDistance = nextDistance
+                step += 1
+        return step
+```
+
+
+
 ### Go
+
 ```Go
 func jump(nums []int) int {
     dp := make([]int, len(nums))
@@ -240,7 +259,71 @@ func min(a, b int) int {
 }
 ```
 
+```go
+// 贪心版本一
+func jump(nums []int) int {
+    n := len(nums)
+    if n == 1 {
+        return 0
+    }
+    cur, next := 0, 0
+    step := 0
+    for i := 0; i < n; i++ {
+        next = max(nums[i]+i, next)
+        if i == cur {
+            if cur != n-1 {
+                step++
+                cur = next
+                if cur >= n-1 {
+                    return step
+                }
+            } else {
+                return step
+            }
+        }
+    }
+    return step
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
+
+```go
+// 贪心版本二
+func jump(nums []int) int {
+    n := len(nums)
+    if n == 1 {
+        return 0
+    }
+    cur, next := 0, 0
+    step := 0
+    for i := 0; i < n-1; i++ {
+        next = max(nums[i]+i, next)
+        if i == cur {
+            cur = next
+            step++
+        }
+    }
+    return step
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
+
+
+
 ### Javascript
+
 ```Javascript
 var jump = function(nums) {
     let curIndex = 0

problems/0055.跳跃游戏.md

@@ -139,7 +139,31 @@ func canJUmp(nums []int) bool {
 }
 ```
 
+```go
+// 贪心
+func canJump(nums []int) bool {
+    cover := 0
+    n := len(nums)-1
+    for i := 0; i <= cover; i++ { // 每次与覆盖值比较
+        cover = max(i+nums[i], cover) //每走一步都将 cover 更新为最大值
+        if cover >= n {
+            return true
+        }
+    }
+    return false
+}
+func max(a, b int ) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```
+
+
+
 ### Javascript 
+
 ```Javascript
 var canJump = function(nums) {
     if(nums.length === 1) return true

problems/0072.编辑距离.md

@@ -362,7 +362,33 @@ function minDistance(word1: string, word2: string): number {
 };
 ```
 
+C：
 
 
+```c
+int min(int num1, int num2, int num3) {
+    return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1); 
+}
+
+int minDistance(char * word1, char * word2){
+    int dp[strlen(word1)+1][strlen(word2)+1];
+    dp[0][0] = 0;
+    for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
+    for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;
+
+    for (int i = 1; i <= strlen(word1); i++) {
+        for (int j = 1; j <= strlen(word2); j++) {
+            if (word1[i-1] == word2[j-1]) {
+                dp[i][j] = dp[i-1][j-1];
+            }
+            else {
+                dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1;
+            }
+        }
+    }
+    return dp[strlen(word1)][strlen(word2)];
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0090.子集II.md

@@ -367,6 +367,39 @@ function subsetsWithDup(nums: number[]): number[][] {
 };
 ```
 
+set去重版本:
+```typescript
+// 使用set去重版本
+function subsetsWithDup(nums: number[]): number[][] {
+    const result: number[][] = [];
+    const path: number[] = [];
+    // 去重之前先排序
+    nums.sort((a, b) => a - b);
+    function backTracking(startIndex: number) {
+        // 收集结果
+        result.push([...path])
+        // 此处不返回也可以因为，每次递归都会使startIndex + 1，当这个数大到nums.length的时候就不会进入递归了。
+        if (startIndex === nums.length) {
+            return
+        }
+        // 定义每一个树层的set集合
+        const set: Set<number> = new Set()
+        for (let i = startIndex; i < nums.length; i++) {
+            // 去重
+            if (set.has(nums[i])) {
+                continue
+            }
+            set.add(nums[i])
+            path.push(nums[i])
+            backTracking(i + 1)
+            // 回溯
+            path.pop()
+        }
+    }
+    backTracking(0)
+    return result
+};
+```
 ### Rust
 
 ```Rust

problems/0104.二叉树的最大深度.md

@@ -313,6 +313,31 @@ class solution {
 }
 ```
 
+```java
+class Solution {
+  /**
+   * 递归法(求深度法)
+   */
+    //定义最大深度
+    int maxnum = 0;
+
+    public int maxDepth(TreeNode root) {
+        ans(root,0);
+        return maxnum;
+    }
+    
+    //递归求解最大深度
+    void ans(TreeNode tr,int tmp){
+        if(tr==null) return;
+        tmp++;
+        maxnum = maxnum<tmp?tmp:maxnum;
+        ans(tr.left,tmp);
+        ans(tr.right,tmp);
+        tmp--;
+    }
+}
+```
+
 ```java
 class solution {
     /**
@@ -552,6 +577,29 @@ func maxdepth(root *treenode) int {
 
 ```
 
+### 559. n叉树的最大深度
+
+```go
+func maxDepth(root *Node) int {
+    if root == nil {
+        return 0
+    }
+    q := list.New()
+    q.PushBack(root)
+    depth := 0
+    for q.Len() > 0 {
+        n := q.Len()
+        for i := 0; i < n; i++ {
+            node := q.Remove(q.Front()).(*Node)
+            for j := range node.Children {
+                q.PushBack(node.Children[j])
+            }
+        }
+        depth++
+    }
+    return depth 
+}
+```
 
 ## javascript 
 

problems/0129.求根到叶子节点数字之和.md

@@ -320,8 +320,11 @@ TypeScript：
 ```typescript
 function sumNumbers(root: TreeNode | null): number {
     if (root === null) return 0;
+    // 记录最终结果
     let resTotal: number = 0;
+    // 记录路径中遇到的节点值
     const route: number[] = [];
+    // 递归初始值
     route.push(root.val);
     recur(root, route);
     return resTotal;
@@ -342,6 +345,7 @@ function sumNumbers(root: TreeNode | null): number {
         };
     }
     function listToSum(nums: number[]): number {
+        // 数组求和
         return Number(nums.join(''));
     }
 };

problems/0139.单词拆分.md

@@ -344,6 +344,21 @@ func wordBreak(s string,wordDict []string) bool  {
 	}
 	return dp[len(s)]
 }
+// 转化为 求装满背包s的前几位字符的方式有几种
+func wordBreak(s string, wordDict []string) bool {
+	// 装满背包s的前几位字符的方式有几种
+	dp := make([]int, len(s)+1)
+	dp[0] = 1
+	for i := 0; i <= len(s); i++ { // 背包
+		for j := 0; j < len(wordDict); j++ { // 物品
+			if i >= len(wordDict[j]) && wordDict[j] == s[i-len(wordDict[j]):i] {
+				dp[i] += dp[i-len(wordDict[j])]
+			}
+		}
+	}
+
+	return dp[len(s)] > 0
+}
 ```
 
 Javascript：

problems/0150.逆波兰表达式求值.md

@@ -89,7 +89,12 @@ C++代码如下：
 class Solution {
 public:
     int evalRPN(vector<string>& tokens) {
+<<<<<<< HEAD
         stack<long long> st;
+=======
+        // 力扣修改了后台测试数据，需要用longlong
+        stack<long long> st; 
+>>>>>>> 28f3b52a82e3cc650290fb02030a53900e122f43
         for (int i = 0; i < tokens.size(); i++) {
             if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
                 long long num1 = st.top();
@@ -104,11 +109,13 @@ public:
                 st.push(stoll(tokens[i]));
             }
         }
+
         int result = st.top();
         st.pop(); // 把栈里最后一个元素弹出（其实不弹出也没事）
         return result;
     }
 };
+
 ```
 
 ## 题外话

problems/0151.翻转字符串里的单词.md

@@ -360,6 +360,75 @@ class Solution {
 }
 ```
 
+```java
+/*
+ * 解法四：时间复杂度 O(n)
+ * 参考卡哥 c++ 代码的三步骤：先移除多余空格，再将整个字符串反转，最后把单词逐个反转
+ * 有别于解法一 ：没有用 StringBuilder  实现，而是对 String 的 char[] 数组操作来实现以上三个步骤
+ */
+class Solution {
+    //用 char[] 来实现 String 的 removeExtraSpaces，reverse 操作
+    public String reverseWords(String s) {
+        char[] chars = s.toCharArray();
+        //1.去除首尾以及中间多余空格
+        chars = removeExtraSpaces(chars);
+        //2.整个字符串反转
+        reverse(chars, 0, chars.length - 1);
+        //3.单词反转
+        reverseEachWord(chars);
+        return new String(chars);
+    }
+
+    //1.用 快慢指针 去除首尾以及中间多余空格，可参考数组元素移除的题解
+    public char[] removeExtraSpaces(char[] chars) {
+        int slow = 0;
+        for (int fast = 0; fast < chars.length; fast++) {
+            //先用 fast 移除所有空格
+            if (chars[fast] != ' ') {
+                //在用 slow 加空格。 除第一个单词外，单词末尾要加空格
+                if (slow != 0)
+                    chars[slow++] = ' ';
+                //fast 遇到空格或遍历到字符串末尾，就证明遍历完一个单词了
+                while (fast < chars.length && chars[fast] != ' ')
+                    chars[slow++] = chars[fast++];
+            }
+        }
+        //相当于 c++ 里的 resize()
+        char[] newChars = new char[slow];
+        System.arraycopy(chars, 0, newChars, 0, slow); 
+        return newChars;
+    }
+
+    //双指针实现指定范围内字符串反转，可参考字符串反转题解
+    public void reverse(char[] chars, int left, int right) {
+        if (right >= chars.length) {
+            System.out.println("set a wrong right");
+            return;
+        }
+        while (left < right) {
+            chars[left] ^= chars[right];
+            chars[right] ^= chars[left];
+            chars[left] ^= chars[right];
+            left++;
+            right--;
+        }
+    }
+
+    //3.单词反转
+    public void reverseEachWord(char[] chars) {
+        int start = 0;
+        //end <= s.length() 这里的 = ，是为了让 end 永远指向单词末尾后一个位置，这样 reverse 的实参更好设置
+        for (int end = 0; end <= chars.length; end++) {
+            // end 每次到单词末尾后的空格或串尾,开始反转单词
+            if (end == chars.length || chars[end] == ' ') {
+                reverse(chars, start, end - 1);
+                start = end + 1;
+            }
+        }
+    }
+}
+```
+
 python:
 
 ```Python

problems/0206.翻转链表.md

@@ -588,5 +588,45 @@ object Solution {
 
 }
 ```
+
+Rust:
+双指针法：
+
+```rust
+impl Solution {
+    pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
+        let mut cur = head;
+        let mut pre = None;
+        while let Some(mut node) = cur.take() {
+            cur = node.next;
+            node.next = pre;
+            pre = Some(node);
+        }
+        pre
+    }
+}
+```
+
+递归法：
+
+```rust
+impl Solution {
+    pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
+        fn rev(
+            mut head: Option<Box<ListNode>>,
+            mut pre: Option<Box<ListNode>>,
+        ) -> Option<Box<ListNode>> {
+            if let Some(mut node) = head.take() {
+                let cur = node.next;
+                node.next = pre;
+                pre = Some(node);
+                return rev(cur, pre);
+            }
+            pre
+        }
+        rev(head, None)
+    }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0209.长度最小的子数组.md

@@ -181,23 +181,23 @@ class Solution:
                 index += 1
         return 0 if res==float("inf") else res
 ```
-```python3
+```python
-#滑动窗口
+# 滑动窗口
 class Solution:
     def minSubArrayLen(self, target: int, nums: List[int]) -> int:
-        if nums is None or len(nums)==0: 
+        if nums is None or len(nums) == 0: 
             return 0 
-        lenf=len(nums)+1
+        lenf = len(nums) + 1
-        total=0
+        total = 0
-        i=j=0
+        i = j = 0
-        while (j<len(nums)):
+        while (j < len(nums)):
-            total=total+nums[j]
+            total = total + nums[j]
-            j+=1
+            j += 1
-            while (total>=target):
+            while (total >= target):
-                lenf=min(lenf,j-i)
+                lenf = min(lenf, j - i)
-                total=total-nums[i]
+                total = total - nums[i]
-                i+=1
+                i += 1
-        if lenf==len(nums)+1:
+        if lenf == len(nums) + 1:
             return 0
         else: 
             return lenf

problems/0222.完全二叉树的节点个数.md

@@ -437,6 +437,33 @@ func countNodes(root *TreeNode) int {
 }
 ```
 
+迭代法
+
+```go
+func countNodes(root *TreeNode) int {
+    if root == nil {
+        return 0
+    }
+    q := list.New()
+    q.PushBack(root)
+    res := 0
+    for q.Len() > 0 {
+        n := q.Len()
+        for i := 0; i < n; i++ {
+            node := q.Remove(q.Front()).(*TreeNode)
+            if node.Left != nil {
+                q.PushBack(node.Left)
+            }
+            if node.Right != nil {
+                q.PushBack(node.Right)
+            }
+            res++
+        }
+    }
+    return res 
+}
+```
+
 
 
 ## JavaScript:

problems/0225.用队列实现栈.md

@@ -292,6 +292,40 @@ class MyStack {
 }
 ```
 
+优化，使用一个 Queue 实现
+```java
+class MyStack {
+
+    Queue<Integer> queue;
+
+    public MyStack() {
+        queue = new LinkedList<>();
+    }
+
+    //每 offer 一个数（A）进来，都重新排列，把这个数（A）放到队列的队首
+    public void push(int x) {
+        queue.offer(x);
+        int size = queue.size();
+        //移动除了 A 的其它数
+        while (size-- > 1)
+            queue.offer(queue.poll());
+    }
+
+    public int pop() {
+        return queue.poll();
+    }
+
+    public int top() {
+        return queue.peek();
+    }
+
+    public boolean empty() {
+        return queue.isEmpty();
+    }
+}
+
+```
+
 Python：
 
 ```python

problems/0455.分发饼干.md

@@ -274,6 +274,7 @@ function findContentChildren(g: number[], s: number[]): number {
 ### C
 
 ```c
+///小餅乾先餵飽小胃口的
 int cmp(int* a, int* b) {
     return *a - *b;
 }
@@ -296,6 +297,33 @@ int findContentChildren(int* g, int gSize, int* s, int sSize){
 }
 ```
 
+```c
+///大餅乾先餵飽大胃口的
+int cmp(int* a, int* b) {
+    return *a - *b;
+}
+
+int findContentChildren(int* g, int gSize, int* s, int sSize){
+    if(sSize == 0)
+        return 0;
+
+    //将两个数组排序为升序
+    qsort(g, gSize, sizeof(int), cmp);
+    qsort(s, sSize, sizeof(int), cmp);
+
+    int count = 0;
+    int start = sSize - 1;
+
+    for(int i = gSize - 1; i >= 0; i--) {
+        if(start >= 0 && s[start] >= g[i] ) {
+            start--;
+            count++;
+        }
+    }
+    return count;
+}
+```
+
 ### Scala
 
 ```scala

problems/0583.两个字符串的删除操作.md

@@ -47,6 +47,8 @@ dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word
 
 那最后当然是取最小值，所以当word1[i - 1] 与 word2[j - 1]不相同的时候，递推公式：dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
 
+因为dp[i - 1][j - 1] + 1等于 dp[i - 1][j] 或 dp[i][j - 1]，所以递推公式可简化为：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
+
 
 3. dp数组如何初始化
 
@@ -90,7 +92,7 @@ public:
                 if (word1[i - 1] == word2[j - 1]) {
                     dp[i][j] = dp[i - 1][j - 1];
                 } else {
-                    dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
+                    dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
                 }
             }
         }

problems/0707.设计链表.md

@@ -1240,6 +1240,94 @@ class MyLinkedList() {
 }
 ```
 
+Rust:
+
+```rust
+#[derive(Debug)]
+pub struct MyLinkedList {
+    pub val: i32,
+    pub next: Option<Box<MyLinkedList>>,
+}
+
+impl MyLinkedList {
+    fn new() -> Self {
+        // 增加头节点
+        MyLinkedList { val: 0, next: None }
+    }
+
+    fn get(&self, index: i32) -> i32 {
+        if index < 0 {
+            return -1;
+        }
+        let mut i = 0;
+        let mut cur = &self.next;
+        while let Some(node) = cur {
+            if i == index {
+                return node.val;
+            }
+            i += 1;
+            cur = &node.next;
+        }
+        -1
+    }
+
+    fn add_at_head(&mut self, val: i32) {
+        let new_node = Box::new(MyLinkedList {
+            val,
+            next: self.next.take(),
+        });
+        self.next = Some(new_node);
+    }
+
+    fn add_at_tail(&mut self, val: i32) {
+        let new_node = Box::new(MyLinkedList { val, next: None });
+        let mut last_node = &mut self.next;
+        while let Some(node) = last_node {
+            last_node = &mut node.next;
+        }
+        *last_node = Some(new_node);
+    }
+
+    fn add_at_index(&mut self, index: i32, val: i32) {
+        if index <= 0 {
+            self.add_at_head(val);
+        } else {
+            let mut i = 0;
+            let mut cur = &mut self.next;
+            while let Some(node) = cur {
+                if i + 1 == index {
+                    let new_node = Box::new(MyLinkedList {
+                        val,
+                        next: node.next.take(),
+                    });
+                    node.next = Some(new_node);
+                    break;
+                }
+                i += 1;
+                cur = &mut node.next;
+            }
+        }
+    }
+
+    fn delete_at_index(&mut self, index: i32) {
+        if index < 0 {
+            return;
+        }
+
+        let mut i = 0;
+        let mut cur = self;
+        while let Some(node) = cur.next.take() {
+            if i == index {
+                cur.next = node.next;
+                break;
+            }
+            i += 1;
+            cur.next = Some(node);
+            cur = cur.next.as_mut().unwrap();
+        }
+    }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/1049.最后一块石头的重量II.md

@@ -43,7 +43,7 @@
 
 是不是感觉和昨天讲解的[416. 分割等和子集](https://programmercarl.com/0416.分割等和子集.html)非常像了。
 
-本题物品的重量为store[i]，物品的价值也为store[i]。
+本题物品的重量为stones[i]，物品的价值也为stones[i]。
 
 对应着01背包里的物品重量weight[i]和 物品价值value[i]。
 

problems/剑指Offer05.替换空格.md

@@ -156,20 +156,20 @@ char* replaceSpace(char* s){
 Java：
 ```Java
 //使用一个新的对象，复制 str，复制的过程对其判断，是空格则替换，否则直接复制，类似于数组复制
-public static String replaceSpace(StringBuffer str) {
+public static String replaceSpace(String s) {
-        if (str == null) {
+        if (s == null) {
             return null;
         }
-		//选用 StringBuilder 单线程使用，比较快，选不选都行
+        //选用 StringBuilder 单线程使用，比较快，选不选都行
         StringBuilder sb = new StringBuilder();
-		//使用 sb 逐个复制 str ，碰到空格则替换，否则直接复制
+        //使用 sb 逐个复制 s ，碰到空格则替换，否则直接复制
-        for (int i = 0; i < str.length(); i++) {
+        for (int i = 0; i < s.length(); i++) {
-		//str.charAt(i) 为 char 类型，为了比较需要将其转为和 " " 相同的字符串类型
+            //s.charAt(i) 为 char 类型，为了比较需要将其转为和 " " 相同的字符串类型
-        //if (" ".equals(String.valueOf(str.charAt(i)))){
+            //if (" ".equals(String.valueOf(s.charAt(i)))){}
             if (s.charAt(i) == ' ') {
                 sb.append("%20");
             } else {
-                sb.append(str.charAt(i));
+                sb.append(s.charAt(i));
             }
         }
         return sb.toString();

problems/剑指Offer58-II.左旋转字符串.md

@@ -115,6 +115,29 @@ class Solution {
 }
 ```
 
+```java
+//解法二：空间复杂度：O(1)。用原始数组来进行反转操作
+//思路为：先整个字符串反转，再反转前面的，最后反转后面 n 个
+class Solution {
+    public String reverseLeftWords(String s, int n) {
+        char[] chars = s.toCharArray();
+        reverse(chars, 0, chars.length - 1);
+        reverse(chars, 0, chars.length - 1 - n);
+        reverse(chars, chars.length - n, chars.length - 1);
+        return new String(chars);
+    }
+
+    public void reverse(char[] chars, int left, int right) {
+        while (left < right) {
+            chars[left] ^= chars[right];
+            chars[right] ^= chars[left];
+            chars[left] ^= chars[right];
+            left++;
+            right--;
+        }
+    }
+```
+
 python: 
 
 ```python

problems/双指针总结.md

@@ -42,7 +42,7 @@ for (int i = 0; i < array.size(); i++) {
 
 **其实很多数组（字符串）填充类的问题，都可以先预先给数组扩容带填充后的大小，然后在从后向前进行操作。** 
 
-那么在[字符串：花式反转还不够！](https://programmercarl.com/0151.翻转字符串里的单词.html)中，我们使用双指针法，用O(n)的时间复杂度完成字符串删除类的操作，因为题目要产出冗余空格。
+那么在[字符串：花式反转还不够！](https://programmercarl.com/0151.翻转字符串里的单词.html)中，我们使用双指针法，用O(n)的时间复杂度完成字符串删除类的操作，因为题目要删除冗余空格。
 
 **在删除冗余空格的过程中，如果不注意代码效率，很容易写成了O(n^2)的时间复杂度。其实使用双指针法O(n)就可以搞定。** 
 

problems/链表理论基础.md

@@ -218,6 +218,22 @@ class ListNode(_x: Int = 0, _next: ListNode = null) {
 }
 ```
 
+Rust:
+```rust
+#[derive(PartialEq, Eq, Clone, Debug)]
+pub struct ListNode<T> {
+    pub val: T,
+    pub next: Option<Box<ListNode<T>>>,
+}
+
+impl<T> ListNode<T> {
+    #[inline]
+    fn new(val: T, node: Option<Box<ListNode<T>>>) -> Self {
+        ListNode { next: node, val }
+    }
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>