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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其实就是判断最常合法长度
如果当前字符为‘(‘ 忽略。
如果为 ‘)‘:
1. 前一字符为 ‘(‘,那么长度是前二字符的最长合法长度+2.
2. 前一字符为‘)‘,那么获得前一字符的最长合法长度tmpLen
a. 如果 前tmpLen+1 项是‘(‘,那么便是合法的,取值为前tmpLen+1+1 项的长度 + tmpLen+2.
b. 为‘)‘,配对失败,忽略。
最后返回最长的。
#include <string> #include <iostream> #include <vector> using namespace std; class Solution { public: int longestValidParentheses(string s) { int len = s.length(); if(len<1) return 0; vector<int > table(len+1,0); int cnt = 0; if(s[0]==‘(‘) cnt++; else cnt --; int retMax = 0; for(int i=1;i<len;i++){ if(s[i]==‘(‘){ if(cnt<0) cnt=1; else cnt++; continue; } cnt--; if(cnt>=0){ if(s[i-1]==‘(‘) table[i+1] = table[i-1]+2; else{ if(s[i-1-table[i]]==‘(‘) table[i+1] = table[i-1-table[i]]+2+table[i]; } if(retMax<table[i+1]) retMax = table[i+1]; } } return retMax; } }; int main() { Solution sol; cout<<sol.longestValidParentheses("()(())")<<endl; return 0; }
时间: 2024-12-29 01:47:56