Merge branch 'master' into master

Author: youngyangyang04

problems/0020.有效的括号.md

@@ -283,6 +283,7 @@ var isValid = function(s) {
 };
 ```
 
+
 Swift
 ```swift
 func isValid(_ s: String) -> Bool {
@@ -307,5 +308,46 @@ func isValid(_ s: String) -> Bool {
 }
 ```
 
+C:
+```C
+//辅助函数：判断栈顶元素与输入的括号是否为一对。若不是，则返回False
+int notMatch(char par, char* stack, int stackTop) {
+    switch(par) {
+        case ']':
+            return stack[stackTop - 1] != '[';
+        case ')':
+            return stack[stackTop - 1] != '(';
+        case '}':
+            return stack[stackTop - 1] != '{';
+    }
+    return 0;
+}
+
+bool isValid(char * s){
+    int strLen = strlen(s);
+    //开辟栈空间
+    char stack[5000];
+    int stackTop = 0;
+
+    //遍历字符串
+    int i;
+    for(i = 0; i < strLen; i++) {
+        //取出当前下标所对应字符
+        char tempChar = s[i];
+        //若当前字符为左括号，则入栈
+        if(tempChar == '(' || tempChar == '[' || tempChar == '{')
+            stack[stackTop++] = tempChar;
+        //若当前字符为右括号，且栈中无元素或右括号与栈顶元素不符，返回False
+        else if(stackTop == 0 || notMatch(tempChar, stack, stackTop))
+            return 0;
+        //当前字符与栈顶元素为一对括号，将栈顶元素出栈
+        else
+            stackTop--;
+    }
+    //若栈中有元素，返回False。若没有元素（stackTop为0），返回True
+    return !stackTop;
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0131.分割回文串.md

@@ -450,6 +450,101 @@ var partition = function(s) {
 };
 ```
 
+##C
+```c
+char** path;
+int pathTop;
+char*** ans;
+int ansTop = 0;
+int* ansSize;
+
+//将path中的字符串全部复制到ans中
+void copy() {
+    //创建一个临时tempPath保存path中的字符串
+    char** tempPath = (char**)malloc(sizeof(char*) * pathTop);
+    int i;
+    for(i = 0; i < pathTop; i++) {
+        tempPath[i] = path[i];
+    }
+    //保存tempPath
+    ans[ansTop] = tempPath;
+    //将当前path的长度（pathTop）保存在ansSize中
+    ansSize[ansTop++] = pathTop;
+}
+
+//判断字符串是否为回文字符串
+bool isPalindrome(char* str, int startIndex, int endIndex) {
+    //双指针法：当endIndex（右指针）的值比startIndex（左指针）大时进行遍历
+    while(endIndex >= startIndex) {
+        //若左指针和右指针指向元素不一样，返回False
+        if(str[endIndex--] != str[startIndex++])
+            return 0;
+    }
+    return 1;
+}
+
+//切割从startIndex到endIndex子字符串
+char* cutString(char* str, int startIndex, int endIndex) {
+    //开辟字符串的空间
+    char* tempString = (char*)malloc(sizeof(char) * (endIndex - startIndex + 2));
+    int i;
+    int index = 0;
+    //复制子字符串
+    for(i = startIndex; i <= endIndex; i++)
+        tempString[index++] = str[i];
+    //用'\0'作为字符串结尾
+    tempString[index] = '\0';
+    return tempString;
+}
+
+void backTracking(char* str, int strLen,  int startIndex) {
+    if(startIndex >= strLen) {
+        //将path拷贝到ans中
+        copy();
+        return ;
+    }
+
+    int i;
+    for(i = startIndex; i < strLen; i++) {
+        //若从subString到i的子串是回文字符串，将其放入path中
+        if(isPalindrome(str, startIndex, i)) {
+            path[pathTop++] = cutString(str, startIndex, i);
+        }
+        //若从startIndex到i的子串不为回文字符串，跳过这一层 
+        else {
+            continue;
+        }
+        //递归判断下一层
+        backTracking(str, strLen, i + 1);
+        //回溯，将path中最后一位元素弹出
+        pathTop--;
+    }
+}
+
+char*** partition(char* s, int* returnSize, int** returnColumnSizes){
+    int strLen = strlen(s);
+    //因为path中的字符串最多为strLen个（即单个字符的回文字符串），所以开辟strLen个char*空间
+    path = (char**)malloc(sizeof(char*) * strLen);
+    //存放path中的数组结果
+    ans = (char***)malloc(sizeof(char**) * 40000);
+    //存放ans数组中每一个char**数组的长度
+    ansSize = (int*)malloc(sizeof(int) * 40000);
+    ansTop = pathTop = 0;
+
+    //回溯函数
+    backTracking(s, strLen, 0);
+
+    //将ansTop设置为ans数组的长度
+    *returnSize = ansTop;
+    //设置ans数组中每一个数组的长度
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int i;
+    for(i = 0; i < ansTop; ++i) {
+        (*returnColumnSizes)[i] = ansSize[i];
+    }
+    return ans;
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>