32. Longest Valid Parentheses
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
原题的意思是,找到原串中最大的括号排列正确的子串的长度。
解决方法: 拿一个max存放目前为止最大的括号串。每次从下一位开始。 重点是怎么寻找从当前位置开始最大的匹配串: 左括号++,右括号++。判断左括号是否大于等于右括号。不满足则错误,退出。且长度为 j - i+1。
很顺利 一遍Accept:
class Solution { public: int longestValidParentheses(string s) { int max = 0; vector<int> num(s.length()); for(int i = 0; i < s.length();i++){ int left = 0,right = 0; for(int j = i; j < s.length(); j++){ if(s[j] == '(') left++; //左括号加一 else right++; if(right > left) break; //当前右括号大于左括号时退出 if(left - right > s.length() - j) break; //需要的右括号数大于要找的剩余位置数,说明不可能凑齐右括号了(这一行可减少时间复杂度) if(right == left) num[i] = j - i + 1; } if(max < num[i]) max = num[i]; } return max; } };