From 52750d3a17c85ba006eed636f2fd33b6089dd9fa Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Thu, 25 Nov 2021 11:59:19 +0000 Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200020.=E6=9C=89?= =?UTF-8?q?=E6=95=88=E7=9A=84=E6=8B=AC=E5=8F=B7.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0020.有效的括号.md | 41 ++++++++++++++++++++++++++++++++ 1 file changed, 41 insertions(+) diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md index 57aa5a01..6c6f0cb7 100644 --- a/problems/0020.有效的括号.md +++ b/problems/0020.有效的括号.md @@ -283,6 +283,47 @@ var isValid = function(s) { }; ``` +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; +} +``` + -----------------------
From 9b6e798285eeb28dcd0ab117c25a75ff03e85c38 Mon Sep 17 00:00:00 2001 From: ArthurP Date: Sat, 27 Nov 2021 19:39:02 +0000 Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200131.=E5=88=86?= =?UTF-8?q?=E5=89=B2=E5=9B=9E=E6=96=87=E4=B8=B2.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 95 ++++++++++++++++++++++++++++++++ 1 file changed, 95 insertions(+) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index f674835d..075734ea 100644 --- a/problems/0131.分割回文串.md +++ b/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; +} +``` -----------------------