Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0027.移除元素.md

@@ -281,10 +281,8 @@ 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
+                slowIndex += 1
         }
     }
     return slowIndex

problems/0045.跳跃游戏II.md

@@ -250,6 +250,27 @@ var jump = function(nums) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function jump(nums: number[]): number {
+    const length: number = nums.length;
+    let curFarthestIndex: number = 0,
+        nextFarthestIndex: number = 0;
+    let curIndex: number = 0;
+    let stepNum: number = 0;
+    while (curIndex < length - 1) {
+        nextFarthestIndex = Math.max(nextFarthestIndex, curIndex + nums[curIndex]);
+        if (curIndex === curFarthestIndex) {
+            curFarthestIndex = nextFarthestIndex;
+            stepNum++;
+        }
+        curIndex++;
+    }
+    return stepNum;
+};
+```
+
 
 
 

problems/0055.跳跃游戏.md

@@ -154,6 +154,23 @@ var canJump = function(nums) {
 };
 ```
 
+### TypeScript
+
+```typescript
+function canJump(nums: number[]): boolean {
+    let farthestIndex: number = 0;
+    let cur: number = 0;
+    while (cur <= farthestIndex) {
+        farthestIndex = Math.max(farthestIndex, cur + nums[cur]);
+        if (farthestIndex >= nums.length - 1) return true;
+        cur++;
+    }
+    return false;
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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 +264,19 @@ 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];
+};
+```
+
+TypeScript：
+
+```typescript
+function maxProfit(prices: number[]): number {
+    let resProfit: number = 0;
+    for (let i = 1, length = prices.length; i < length; i++) {
+        resProfit += Math.max(prices[i] - prices[i - 1], 0);
+    }
+    return resProfit;
 };
 ```
 

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

@@ -235,7 +235,7 @@ class Solution {
         return index;
     }
 }
-``` 
+```
 
 ### Python 
 ```python
@@ -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

@@ -238,6 +238,32 @@ var candy = function(ratings) {
 };
 ```
 
+### 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/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/0707.设计链表.md

@@ -973,6 +973,10 @@ class MyLinkedList {
         // 处理头节点
         if (index === 0) {
             this.head = this.head!.next;
+            // 如果链表中只有一个元素，删除头节点后，需要处理尾节点
+            if (index === this.size - 1) {
+                this.tail = null
+            }
             this.size--;
             return;
         }

problems/0763.划分字母区间.md

@@ -77,6 +77,61 @@ public:
 
 但这道题目的思路是很巧妙的，所以有必要介绍给大家做一做，感受一下。
 
+## 补充
+
+这里提供一种与[452.用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)、[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)相同的思路。
+
+统计字符串中所有字符的起始和结束位置，记录这些区间(实际上也就是[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)题目里的输入)，**将区间按左边界从小到大排序，找到边界将区间划分成组，互不重叠。找到的边界就是答案。**
+
+```CPP
+class Solution {
+public:
+    static bool cmp(vector<int> &a, vector<int> &b) {
+        return a[0] < b[0];
+    }
+    // 记录每个字母出现的区间
+    vector<vector<int>> countLabels(string s) {
+        vector<vector<int>> hash(26, vector<int>(2, INT_MIN));
+        vector<vector<int>> hash_filter;
+        for (int i = 0; i < s.size(); ++i) {
+            if (hash[s[i] - 'a'][0] == INT_MIN) {
+                hash[s[i] - 'a'][0] = i;
+            }
+            hash[s[i] - 'a'][1] = i;
+        }
+        // 去除字符串中未出现的字母所占用区间
+        for (int i = 0; i < hash.size(); ++i) {
+            if (hash[i][0] != INT_MIN) {
+                hash_filter.push_back(hash[i]);
+            }
+        }
+        return hash_filter;
+    }
+    vector<int> partitionLabels(string s) {
+        vector<int> res;
+        // 这一步得到的 hash 即为无重叠区间题意中的输入样例格式：区间列表
+        // 只不过现在我们要求的是区间分割点
+        vector<vector<int>> hash = countLabels(s);
+        // 按照左边界从小到大排序
+        sort(hash.begin(), hash.end(), cmp);
+        // 记录最大右边界
+        int rightBoard = hash[0][1];
+        int leftBoard = 0;
+        for (int i = 1; i < hash.size(); ++i) {
+            // 由于字符串一定能分割，因此,
+            // 一旦下一区间左边界大于当前右边界，即可认为出现分割点
+            if (hash[i][0] > rightBoard) {
+                res.push_back(rightBoard - leftBoard + 1);
+                leftBoard = hash[i][0];
+            }
+            rightBoard = max(rightBoard, hash[i][1]);
+        }
+        // 最右端
+        res.push_back(rightBoard - leftBoard + 1);
+        return res;
+    }
+};
+```
 
 ## 其他语言版本
 

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 +211,29 @@ var largestSumAfterKNegations = function(nums, k) {
 };
 ```
 
+### 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/前序/ACM模式如何构建二叉树.md

@@ -208,6 +208,59 @@ int main() {
 ## Java 
 
 ```Java
+public class Solution {
+    // 节点类
+    static class TreeNode {
+        // 节点值
+        int val;
+
+        // 左节点
+        TreeNode left;
+
+        // 右节点
+        TreeNode right;
+
+        // 节点的构造函数(默认左右节点都为null)
+        public TreeNode(int x) {
+            this.val = x;
+            this.left = null;
+            this.right = null;
+        }
+    }
+    
+    /**
+     * 根据数组构建二叉树
+     * @param arr 树的数组表示
+     * @return 构建成功后树的根节点
+     */
+    public TreeNode constructBinaryTree(final int[] arr) {
+        // 构建和原数组相同的树节点列表
+        List<TreeNode> treeNodeList = arr.length > 0 ? new ArrayList<>(arr.length) : null;
+        TreeNode root = null;
+        // 把输入数值数组，先转化为二叉树节点列表
+        for (int i = 0; i < arr.length; i++) {
+            TreeNode node = null;
+            if (arr[i] != -1) { // 用 -1 表示null
+                node = new TreeNode(arr[i]);
+            }
+            treeNodeList.add(node);
+            if (i == 0) {
+                root = node;
+            }
+        }
+        // 遍历一遍，根据规则左右孩子赋值就可以了
+        // 注意这里 结束规则是 i * 2 + 2 < arr.length，避免空指针
+        for (int i = 0; i * 2 + 2 < arr.length; i++) {
+            TreeNode node = treeNodeList.get(i);
+            if (node != null) {
+                // 线性存储转连式存储关键逻辑
+                node.left = treeNodeList.get(2 * i + 1);
+                node.right = treeNodeList.get(2 * i + 2);
+            }
+        }
+        return root;
+    }
+}
 ```