Merge branch 'master' into dev

problems/0019.删除链表的倒数第N个节点.md

@@ -338,5 +338,28 @@ 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/0072.编辑距离.md

@@ -364,6 +364,7 @@ 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); 

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/0150.逆波兰表达式求值.md

@@ -89,26 +89,29 @@ C++代码如下：
 class Solution {
 public:
     int evalRPN(vector<string>& tokens) {
-        stack<int> st;
+        // 力扣修改了后台测试数据，需要用longlong
+        stack<long long> st; 
         for (int i = 0; i < tokens.size(); i++) {
             if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {
-                int num1 = st.top();
+                long long num1 = st.top();
                 st.pop();
-                int num2 = st.top();
+                long long num2 = st.top();
                 st.pop();
                 if (tokens[i] == "+") st.push(num2 + num1);
                 if (tokens[i] == "-") st.push(num2 - num1);
                 if (tokens[i] == "*") st.push((long)num2 * (long)num1); //力扣改了后台测试数据
                 if (tokens[i] == "/") st.push(num2 / num1);
             } else {
-                st.push(stoi(tokens[i]));
+                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/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/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>