leetcode 32 -- Longest Valid Parentheses

Longest Valid Parentheses

题目：
Given a string containing just the characters ‘(’ and ‘)’, find the length of the longest valid (well-formed) parentheses substring.
For “(()”, the longest valid parentheses substring is “()”, which has length = 2.
Another example is “)()())”, where the longest valid parentheses substring is “()()”, which has length = 4.

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题意：
给定一个字符串，找出其中满足括号匹配的最长字串的长度。

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思路：
1.类似暴力求解，对每个子串进行验证，记录其括号匹配的长度，括号匹配原则，任意时刻，左括号个数一定大于右括号个数，且起始不能为右括号。时间复杂度O(n^2)，超时。
2.新建一个数组（长度和s相同）作为辅助记录字符串s的括号匹配过程，用栈来帮我们测试是否满足括号匹配，有满足匹配则把对应下标置为1，不满足为0，最后遍历辅助数组，最长连续为1的长度就是我们所需要的结果，时间复杂度为O(n)，AC

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代码：

思路1

int longestValidParentheses(char* s) {    char *p = s;    char *q;    int ret = 0;    int left;    int right;    while(*p != '\0'){        //首字符不能为右括号        if(*p == ')'){            ++p;            continue;        }else if(*p == '('){            left = 1;            right = 0;            q = p+1;            while(*q != '\0'){                //printf("s:%s\n", q);                //sleep(1);                if(*q == '('){                    left++;                }else if(*q == ')'){                    right++;                }                //满足匹配更新，不满足跳出循环                if(right == left){                    if(right+left > ret){                        ret = right+left;                        //printf("ret:%d\n", ret);                    }                }else if(right > left){                    p = q;                    ++q;                    break;                }                ++q;            }            ++p;        }    }    return ret;}

手动测试过正确，但是leetcode上面显示超时

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思路2

class Solution {public:    int longestValidParentheses(string s) {        stack<int>sk;        vector<int>ivec(s.length());        //初始化        for(int &i : ivec){            i = 0;        }        for(int i = 0; i < s.length(); ++i){            //注意栈保存的是对应字符的下标            if(s[i] == '('){                sk.push(i);            }else if(s[i] == ')' && !sk.empty()){                //当为')'时，栈若不为空，则里面一定为'('，满足匹配，标记为1                ivec[i] = 1;                ivec[sk.top()] = 1;                sk.pop();            }        }        int ret= 0;        int tmp = 0;        //找最长连续为1的        for(const int &i : ivec){            if(i == 1){                tmp++;            }else{                tmp = 0;            }            if(ret < tmp){                ret = tmp;            }                    }        return ret;    }};