Merge branch 'youngyangyang04:master' into master

Author: gaoyangu

problems/0028.实现strStr.md

@@ -1059,5 +1059,112 @@ func getNext(_ next: inout [Int], needle: [Character]) {
 
 ```
 
+> 前缀表右移
+
+```swift
+func strStr(_ haystack: String, _ needle: String) -> Int {
+        
+        let s = Array(haystack), p = Array(needle)
+        guard p.count != 0 else { return 0 }
+        
+        var j = 0
+        var next = [Int].init(repeating: 0, count: p.count)
+        getNext(&next, p)
+         
+        for i in 0 ..< s.count {
+            
+            while j > 0 && s[i] != p[j] {
+                j = next[j]
+            }
+            
+            if s[i] == p[j] {
+                j += 1
+            }
+            
+            if j == p.count {
+                return i - p.count + 1
+            }
+        }
+        
+        return -1
+    }
+    
+    // 前缀表后移一位，首位用 -1 填充
+    func getNext(_ next: inout [Int], _ needle: [Character]) {
+        
+        guard needle.count > 1 else { return }
+
+        var j = 0
+        next[0] = j
+        
+        for i in 1 ..< needle.count-1 {
+           
+            while j > 0 && needle[i] != needle[j] {
+                j = next[j-1]
+            }
+            
+            if needle[i] == needle[j] {
+                j += 1
+            }
+            
+            next[i] = j
+        }
+        next.removeLast()
+        next.insert(-1, at: 0)
+    }
+```
+
+> 前缀表统一不减一
+```swift
+
+func strStr(_ haystack: String, _ needle: String) -> Int {
+        
+        let s = Array(haystack), p = Array(needle)
+        guard p.count != 0 else { return 0 }
+    
+        var j = 0
+        var next = [Int](repeating: 0, count: needle.count)
+        // KMP
+        getNext(&next, needle: p)
+        
+        for i in 0 ..< s.count {
+            while j > 0 && s[i] != p[j] {
+                j = next[j-1]
+            }
+
+            if s[i] == p[j] {
+                j += 1
+            }
+
+            if j == p.count {
+                return i - p.count + 1
+            }
+        }
+        return -1
+    }
+
+    //前缀表
+    func getNext(_ next: inout [Int], needle: [Character]) {
+       
+        var j = 0
+        next[0] = j
+        
+        for i in 1 ..< needle.count {
+            
+            while j>0 && needle[i] != needle[j] {
+                j = next[j-1]
+            }
+            
+            if needle[i] == needle[j] {
+                j += 1
+            }
+            
+            next[i] = j
+            
+        }
+    }
+
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0053.最大子序和.md

@@ -230,6 +230,60 @@ 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**
@@ -267,5 +321,6 @@ 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/0093.复原IP地址.md

@@ -227,7 +227,7 @@ private:
 public:
     vector<string> restoreIpAddresses(string s) {
         result.clear();
-        if (s.size() > 12) return result; // 算是剪枝了
+        if (s.size() < 4 || s.size() > 12) return result; // 算是剪枝了
         backtracking(s, 0, 0);
         return result;
     }

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -103,7 +103,7 @@ TreeNode* traversal (vector<int>& inorder, vector<int>& postorder) {
 中序数组相对比较好切，找到切割点（后序数组的最后一个元素）在中序数组的位置，然后切割，如下代码中我坚持左闭右开的原则：
 
 
-```C++
+```CPP
 // 找到中序遍历的切割点
 int delimiterIndex;
 for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) {
@@ -130,7 +130,7 @@ vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() );
 
 代码如下：
 
-```
+```CPP
 // postorder 舍弃末尾元素，因为这个元素就是中间节点，已经用过了
 postorder.resize(postorder.size() - 1);
 
@@ -144,7 +144,7 @@ vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end
 
 接下来可以递归了，代码如下：
 
-```
+```CPP
 root->left = traversal(leftInorder, leftPostorder);
 root->right = traversal(rightInorder, rightPostorder);
 ```

problems/0122.买卖股票的最佳时机II.md

@@ -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/0135.分发糖果.md

@@ -238,6 +238,49 @@ 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
@@ -264,6 +307,5 @@ function candy(ratings: 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/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