Merge branch 'youngyangyang04:master' into master

README.md

@@ -531,7 +531,8 @@
 
 如果是已工作，备注：姓名-城市-岗位-组队刷题。如果学生，备注：姓名-学校-年级-组队刷题。**备注没有自我介绍不通过哦**
 
-<div align="center"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20220426233122.png" data-img="1" width="200" height="200"></img></div>
+
+<div align="center"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/第二企业刷题活码.png" data-img="1" width="200" height="200"></img></div>
 
 
 
@@ -543,6 +544,7 @@
 
 **来看看就知道了，你会发现相见恨晚！**
 
+
 <a name="公众号"></a>
 <div align="center"><img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20211026122841.png" data-img="1" width="650" height="500"></img></div>
 

problems/0027.移除元素.md

@@ -281,9 +281,7 @@ func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
 
     for fastIndex in 0..<nums.count {
         if val != nums[fastIndex] {
-            if slowIndex != fastIndex {
                 nums[slowIndex] = nums[fastIndex]
-            }
                 slowIndex += 1
         }
     }

problems/0053.最大子序和.md

@@ -244,6 +244,60 @@ var maxSubArray = function(nums) {
 };
 ```
 
+
+### C:
+贪心：
+```c
+int maxSubArray(int* nums, int numsSize){
+    int maxVal = INT_MIN;
+    int subArrSum = 0;
+    
+    int i;
+    for(i = 0; i < numsSize; ++i) {
+        subArrSum += nums[i];
+        // 若当前局部和大于之前的最大结果，对结果进行更新
+        maxVal = subArrSum > maxVal ? subArrSum : maxVal;
+        // 若当前局部和为负，对结果无益。则从nums[i+1]开始应重新计算。
+        subArrSum = subArrSum < 0 ? 0 : subArrSum;
+    }
+
+    return maxVal;
+}
+```
+
+动态规划：
+```c
+/**
+ * 解题思路：动态规划：
+ * 1. dp数组：dp[i]表示从0到i的子序列中最大序列和的值
+ * 2. 递推公式：dp[i] = max(dp[i-1] + nums[i], nums[i])
+        若dp[i-1]<0，对最后结果无益。dp[i]则为nums[i]。
+ * 3. dp数组初始化：dp[0]的最大子数组和为nums[0]
+ * 4. 推导顺序：从前往后遍历
+ */
+
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+int maxSubArray(int* nums, int numsSize){
+    int dp[numsSize];
+    // dp[0]最大子数组和为nums[0]
+    dp[0] = nums[0];
+    // 若numsSize为1，应直接返回nums[0]
+    int subArrSum = nums[0];
+
+    int i;
+    for(i = 1; i < numsSize; ++i) {
+        dp[i] = max(dp[i - 1] + nums[i], nums[i]);
+
+        // 若dp[i]大于之前记录的最大值，进行更新
+        if(dp[i] > subArrSum)
+            subArrSum = dp[i];
+    }
+
+    return subArrSum;
+}
+```
+
 ### TypeScript
 
 **贪心**
@@ -281,5 +335,6 @@ function maxSubArray(nums: number[]): number {
 
 
 
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0055.跳跃游戏.md

@@ -154,6 +154,30 @@ var canJump = function(nums) {
 };
 ```
 
+
+### C
+```c
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+bool canJump(int* nums, int numsSize){
+    int cover = 0;
+
+    int i;
+    // 只可能获取cover范围中的步数，所以i<=cover
+    for(i = 0; i <= cover; ++i) {
+        // 更新cover为从i出发能到达的最大值/cover的值中较大值
+        cover = max(i + nums[i], cover);
+
+        // 若更新后cover可以到达最后的元素，返回true
+        if(cover >= numsSize - 1)
+            return true;
+    }
+
+    return false;
+}
+```
+
+
 ### TypeScript
 
 ```typescript

problems/0059.螺旋矩阵II.md

@@ -246,11 +246,11 @@ var generateMatrix = function(n) {
             res[row][col] = count++;
         }
         // 下行从右到左（左闭右开）
-        for (; col > startX; col--) {
+        for (; col > startY; col--) {
             res[row][col] = count++;
         }
         // 左列做下到上（左闭右开）
-        for (; row > startY; row--) {
+        for (; row > startX; row--) {
             res[row][col] = count++;
         }
 

problems/0121.买卖股票的最佳时机.md

@@ -311,7 +311,36 @@ class Solution:
 ```
 
 Go:
+> 贪心法：
+```Go
+func maxProfit(prices []int) int {
+    low := math.MaxInt32
+    rlt := 0
+    for i := range prices{
+        low = min(low, prices[i])
+        rlt = max(rlt, prices[i]-low)
+    }
 
+    return rlt
+}
+func min(a, b int) int {
+    if a < b{
+        return a
+    }
+
+    return b
+}
+
+func max(a, b int) int {
+    if a > b{
+        return a
+    }
+
+    return b
+}
+```
+
+> 动态规划：版本一
 ```Go
 func maxProfit(prices []int) int {
 	length:=len(prices)
@@ -338,6 +367,29 @@ func max(a,b int)int {
 }
 ```
 
+> 动态规划：版本二
+```Go
+func maxProfit(prices []int) int {
+    dp := [2][2]int{}
+    dp[0][0] = -prices[0]   
+    dp[0][1] = 0
+    for i := 1; i < len(prices); i++{
+        dp[i%2][0] = max(dp[(i-1)%2][0], -prices[i])
+        dp[i%2][1] = max(dp[(i-1)%2][1], dp[(i-1)%2][0]+prices[i])
+    }
+
+    return dp[(len(prices)-1)%2][1]
+}
+
+func max(a, b int) int {
+    if a > b{
+        return a 
+    }
+
+    return b
+}
+```
+
 JavaScript:
 
 > 动态规划

problems/0122.买卖股票的最佳时机II.md

@@ -264,7 +264,7 @@ const maxProfit = (prices) => {
         dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] + prices[i]);
     }
 
-    return dp[prices.length -1][0];
+    return dp[prices.length -1][1];
 };
 ```
 
@@ -281,7 +281,7 @@ function maxProfit(prices: number[]): number {
 ```
 
 C:
-
+贪心：
 ```c
 int maxProfit(int* prices, int pricesSize){
     int result = 0;
@@ -296,5 +296,27 @@ int maxProfit(int* prices, int pricesSize){
 }
 ```
 
+动态规划：
+```c
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+int maxProfit(int* prices, int pricesSize){
+    int dp[pricesSize][2];
+    dp[0][0] = 0 - prices[0];
+    dp[0][1] = 0;
+
+    int i;
+    for(i = 1; i < pricesSize; ++i) {
+        // dp[i][0]为i-1天持股的钱数/在第i天用i-1天的钱买入的最大值。
+        // 若i-1天持股，且第i天买入股票比i-1天持股时更亏，说明应在i-1天时持股
+        dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i]);
+        //dp[i][1]为i-1天不持股钱数/在第i天卖出所持股票dp[i-1][0] + prices[i]的最大值
+        dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i]);
+    }
+    // 返回在最后一天不持股时的钱数（将股票卖出后钱最大化）
+    return dp[pricesSize - 1][1];
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0122.买卖股票的最佳时机II（动态规划）.md

@@ -276,7 +276,7 @@ const maxProfit = (prices) => {
         dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] + prices[i]);
     }
 
-    return dp[prices.length -1][0];
+    return dp[prices.length -1][1];
 };
 
 // 方法二：动态规划（滚动数组）

problems/0134.加油站.md

@@ -364,7 +364,50 @@ var canCompleteCircuit = function(gas, cost) {
 };
 ```
 
+### TypeScript
+
+**暴力法：**
+
+```typescript
+function canCompleteCircuit(gas: number[], cost: number[]): number {
+    for (let i = 0, length = gas.length; i < length; i++) {
+        let curSum: number = 0;
+        let index: number = i;
+        while (curSum >= 0 && index < i + length) {
+            let tempIndex: number = index % length;
+            curSum += gas[tempIndex] - cost[tempIndex];
+            index++;
+        }
+        if (index === i + length && curSum >= 0) return i;
+    }
+    return -1;
+};
+```
+
+**解法二：**
+
+```typescript
+function canCompleteCircuit(gas: number[], cost: number[]): number {
+    let total: number = 0;
+    let curGas: number = 0;
+    let tempDiff: number = 0;
+    let resIndex: number = 0;
+    for (let i = 0, length = gas.length; i < length; i++) {
+        tempDiff = gas[i] - cost[i];
+        total += tempDiff;
+        curGas += tempDiff;
+        if (curGas < 0) {
+            resIndex = i + 1;
+            curGas = 0;
+        }
+    }
+    if (total < 0) return -1;
+    return resIndex;
+};
+```
+
 ### C 
+
 ```c
 int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
     int curSum = 0;

problems/0135.分发糖果.md

@@ -258,5 +258,73 @@ var candy = function(ratings) {
 ```
 
 
+### C
+```c
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+int *initCandyArr(int size) {
+    int *candyArr = (int*)malloc(sizeof(int) * size);
+
+    int i;
+    for(i = 0; i < size; ++i)
+        candyArr[i] = 1;
+
+    return candyArr;
+}
+
+int candy(int* ratings, int ratingsSize){
+    // 初始化数组，每个小孩开始至少有一颗糖
+    int *candyArr = initCandyArr(ratingsSize);
+
+    int i;
+    // 先判断右边是否比左边评分高。若是，右边孩子的糖果为左边孩子+1（candyArr[i] = candyArr[i - 1] + 1)
+    for(i = 1; i < ratingsSize; ++i) {
+        if(ratings[i] > ratings[i - 1])
+            candyArr[i] = candyArr[i - 1] + 1;
+    }
+
+    // 再判断左边评分是否比右边高。
+    // 若是，左边孩子糖果为右边孩子糖果+1/自己所持糖果最大值。（若糖果已经比右孩子+1多，则不需要更多糖果）
+    // 举例：ratings为[1, 2, 3, 1]。此时评分为3的孩子在判断右边比左边大后为3，虽然它比最末尾的1(ratings[3])大，但是candyArr[3]为1。所以不必更新candyArr[2]
+    for(i = ratingsSize - 2; i >= 0; --i) {
+        if(ratings[i] > ratings[i + 1])
+            candyArr[i] = max(candyArr[i], candyArr[i + 1] + 1);
+    }
+
+    // 求出糖果之和
+    int result = 0;
+    for(i = 0; i < ratingsSize; ++i) {
+        result += candyArr[i];
+    }
+    return result;
+}
+```
+
+### TypeScript
+
+```typescript
+function candy(ratings: number[]): number {
+    const candies: number[] = [];
+    candies[0] = 1;
+    // 保证右边高分孩子一定比左边低分孩子发更多的糖果
+    for (let i = 1, length = ratings.length; i < length; i++) {
+        if (ratings[i] > ratings[i - 1]) {
+            candies[i] = candies[i - 1] + 1;
+        } else {
+            candies[i] = 1;
+        }
+    }
+    // 保证左边高分孩子一定比右边低分孩子发更多的糖果
+    for (let i = ratings.length - 2; i >= 0; i--) {
+        if (ratings[i] > ratings[i + 1]) {
+            candies[i] = Math.max(candies[i], candies[i + 1] + 1);
+        }
+    }
+    return candies.reduce((pre, cur) => pre + cur);
+};
+```
+
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0143.重排链表.md

@@ -336,7 +336,33 @@ class Solution:
         return pre
 ```
 ### Go 
-
+```go
+# 方法三 分割链表
+func reorderList(head *ListNode)  {
+    var slow=head
+    var fast=head
+    for fast!=nil&&fast.Next!=nil{
+        slow=slow.Next
+        fast=fast.Next.Next
+    }  //双指针将链表分为左右两部分
+    var right =new(ListNode)
+    for slow!=nil{
+        temp:=slow.Next
+        slow.Next=right.Next
+        right.Next=slow
+        slow=temp
+    }  //翻转链表右半部分
+    right=right.Next  //right为反转后得右半部分
+    h:=head
+    for right.Next!=nil{
+        temp:=right.Next
+        right.Next=h.Next
+        h.Next=right
+        h=h.Next.Next
+        right=temp
+    } //将左右两部分重新组合
+}
+```
 ### JavaScript 
 
 ```javascript

problems/0376.摆动序列.md

@@ -298,6 +298,35 @@ var wiggleMaxLength = function(nums) {
 };
 ```
 
+
+### C
+**贪心**
+```c
+int wiggleMaxLength(int* nums, int numsSize){
+    if(numsSize <= 1) 
+        return numsSize;
+
+    int length = 1;
+    int preDiff , curDiff;
+    preDiff = curDiff = 0;
+    for(int i = 0; i < numsSize - 1; ++i) {
+        // 计算当前i元素与i+1元素差值
+        curDiff = nums[i+1] - nums[i];
+
+        // 若preDiff与curDiff符号不符，则子序列长度+1。更新preDiff的符号
+        // 若preDiff与curDiff符号一致，当前i元素为连续升序/连续降序子序列的中间元素。不被记录入长度
+        // 注：当preDiff为0时，curDiff为正或为负都属于符号不同
+        if((curDiff > 0 && preDiff <= 0) || (preDiff >= 0 && curDiff < 0)) {
+            preDiff = curDiff;
+            length++;
+        }
+    }
+
+    return length;
+}
+```
+
+
 ### TypeScript
 
 **贪心**

problems/0406.根据身高重建队列.md

@@ -290,6 +290,24 @@ var reconstructQueue = function(people) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function reconstructQueue(people: number[][]): number[][] {
+    people.sort((a, b) => {
+        if (a[0] === b[0]) return a[1] - b[1];
+        return b[0] - a[0];
+    });
+    const resArr: number[][] = [];
+    for (let i = 0, length = people.length; i < length; i++) {
+        resArr.splice(people[i][1], 0, people[i]);
+    }
+    return resArr;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0416.分割等和子集.md

@@ -50,7 +50,7 @@
 
 ## 01背包问题
 
-背包问题，大家都知道，有N件物品和一个最多能被重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。每件物品只能用一次，求解将哪些物品装入背包里物品价值总和最大。
+背包问题，大家都知道，有N件物品和一个最多能背重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。每件物品只能用一次，求解将哪些物品装入背包里物品价值总和最大。
 
 **背包问题有多种背包方式，常见的有：01背包、完全背包、多重背包、分组背包和混合背包等等。**
 

problems/0503.下一个更大元素II.md

@@ -166,20 +166,19 @@ JavaScript:
  * @return {number[]}
  */
 var nextGreaterElements = function (nums) {
-  // let map = new Map();
+  const len = nums.length;
   let stack = [];
-  let res = new Array(nums.length).fill(-1);
+  let res = Array(len).fill(-1);
-  for (let i = 0; i < nums.length * 2; i++) {
+  for (let i = 0; i < len * 2; i++) {
     while (
       stack.length &&
-      nums[i % nums.length] > nums[stack[stack.length - 1]]
+      nums[i % len] > nums[stack[stack.length - 1]]
     ) {
-      let index = stack.pop();
+      const index = stack.pop();
-      res[index] = nums[i % nums.length];
+      res[index] = nums[i % len];
     }
-    stack.push(i % nums.length);
+    stack.push(i % len);
   }
-
   return res;
 };
 ```

problems/0739.每日温度.md

@@ -301,25 +301,22 @@ func dailyTemperatures(num []int) []int {
 
 JavaScript：
 ```javascript
-/**
+// 版本一
- * @param {number[]} temperatures
- * @return {number[]}
- */
 var dailyTemperatures = function(temperatures) {
-    let n = temperatures.length;
+    const n = temperatures.length;
-    let res = new Array(n).fill(0);
+    const res = Array(n).fill(0);
-    let stack = [];  // 递减栈：用于存储元素右面第一个比他大的元素下标
+    const stack = [];  // 递增栈：用于存储元素右面第一个比他大的元素下标
     stack.push(0);
     for (let i = 1; i < n; i++) {
         // 栈顶元素
-        let top = stack[stack.length - 1];
+        const top = stack[stack.length - 1];
         if (temperatures[i] < temperatures[top]) {
             stack.push(i);
         } else if (temperatures[i] === temperatures[top]) {
             stack.push(i);
         } else {
             while (stack.length && temperatures[i] > temperatures[stack[stack.length - 1]]) {
-                let top = stack.pop();
+                const top = stack.pop();
                 res[top] = i - top;
             }
             stack.push(i);
@@ -327,6 +324,23 @@ var dailyTemperatures = function(temperatures) {
     }
     return res;
 };
+
+
+// 版本二
+var dailyTemperatures = function(temperatures) {
+    const n = temperatures.length;
+    const res = Array(n).fill(0);
+    const stack = [];  // 递增栈：用于存储元素右面第一个比他大的元素下标
+    stack.push(0);
+    for (let i = 1; i < n; i++) {
+        while (stack.length && temperatures[i] > temperatures[stack[stack.length - 1]]) {
+            const top = stack.pop();
+            res[top] = i - top;
+        }
+        stack.push(i);
+    }
+    return res;
+};
 ```
 
 

problems/0860.柠檬水找零.md

@@ -252,5 +252,39 @@ var lemonadeChange = function(bills) {
 
 ```
 
+### TypeScript
+
+```typescript
+function lemonadeChange(bills: number[]): boolean {
+    let five: number = 0,
+        ten: number = 0;
+    for (let bill of bills) {
+        switch (bill) {
+            case 5:
+                five++;
+                break;
+            case 10:
+                if (five < 1) return false;
+                five--;
+                ten++
+                break;
+            case 20:
+                if (ten > 0 && five > 0) {
+                    five--;
+                    ten--;
+                } else if (five > 2) {
+                    five -= 3;
+                } else {
+                    return false;
+                }
+                break;
+        }
+    }
+    return true;
+};
+```
+
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/1005.K次取反后最大化的数组和.md

@@ -211,5 +211,70 @@ var largestSumAfterKNegations = function(nums, k) {
 };
 ```
 
+
+### C
+```c
+#define abs(a) (((a) > 0) ? (a) : (-(a)))
+
+// 对数组求和
+int sum(int *nums, int numsSize) {
+    int sum = 0;
+
+    int i;
+    for(i = 0; i < numsSize; ++i) {
+        sum += nums[i];
+    }
+    return sum;
+}
+
+int cmp(const void* v1, const void* v2) {
+    return abs(*(int*)v2) - abs(*(int*)v1);
+}
+
+int largestSumAfterKNegations(int* nums, int numsSize, int k){
+    qsort(nums, numsSize, sizeof(int), cmp);
+
+    int i;
+    for(i = 0; i < numsSize; ++i) {
+        // 遍历数组，若当前元素<0则将当前元素转变，k--
+        if(nums[i] < 0 && k > 0) {
+            nums[i] *= -1;
+            --k;
+        }
+    }
+
+    // 若遍历完数组后k还有剩余（此时所有元素应均为正），则将绝对值最小的元素nums[numsSize - 1]变为负
+    if(k % 2 == 1)
+        nums[numsSize - 1] *= -1;
+
+    return sum(nums, numsSize);
+}
+```
+
+### TypeScript
+
+```typescript
+function largestSumAfterKNegations(nums: number[], k: number): number {
+    nums.sort((a, b) => Math.abs(b) - Math.abs(a));
+    let curIndex: number = 0;
+    const length = nums.length;
+    while (curIndex < length && k > 0) {
+        if (nums[curIndex] < 0) {
+            nums[curIndex] *= -1;
+            k--;
+        }
+        curIndex++;
+    }
+    while (k > 0) {
+        nums[length - 1] *= -1;
+        k--;
+    }
+    return nums.reduce((pre, cur) => pre + cur, 0);
+};
+```
+
+
+
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/1047.删除字符串中的所有相邻重复项.md

@@ -250,11 +250,9 @@ func removeDuplicates(s string) string {
 
 javaScript: 
 
+法一：使用栈
+
 ```js
-/**
- * @param {string} s
- * @return {string}
- */
 var removeDuplicates = function(s) {
     const stack = [];
     for(const x of s) {
@@ -267,6 +265,25 @@ var removeDuplicates = function(s) {
 };
 ```
 
+法二：双指针（模拟栈）
+
+```js
+// 原地解法（双指针模拟栈）
+var removeDuplicates = function(s) {
+    s = [...s];
+    let top = -1; // 指向栈顶元素的下标
+    for(let i = 0; i < s.length; i++) {
+        if(top === -1 || s[top] !== s[i]) { // top === -1 即空栈
+            s[++top] = s[i]; // 入栈
+        } else {
+            top--; // 推出栈
+        }
+    }
+    s.length = top + 1; // 栈顶元素下标 + 1 为栈的长度
+    return s.join('');
+};
+```
+
 TypeScript：
 
 ```typescript