leetcode 32. Longest Valid Parentheses

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于 2016-03-27 11:35:10 发布

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分类专栏：数据结构与算法 OJ 文章标签： leetcode

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本文介绍了一种解决最长有效括号子串问题的方法。通过遍历字符串并将字符及其位置存储为元组，利用栈操作来寻找匹配的括号并删除已匹配的括号对，最终找出最长的有效括号子串长度。

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.

class Solution {
public:
	int longestValidParentheses(string s) {
		vector<pair<char, int>>str;
		for (int i = 0; i < s.length(); i++)
			str.push_back(pair<char, int>(s[i], i));
		bool f = true;
		while (!str.empty()&&f)
		{
			f = false;
			int k = 0;
			while (!str.empty()&&k < str.size()-1)
			{
				if (str[k].first == '('&&str[k + 1].first == ')')
				{
					f = true;
					str.erase(str.begin() + k, str.begin() + k + 2);
				}
				else k++;
			}
		}
		if (str.empty())
			return s.length();
		int maxlen = 0;
		if (str[0].second>maxlen)
			maxlen = str[0].second;
		if (s.length() - str.back().second - 1 > maxlen)
			maxlen = s.length() - str.back().second - 1;
		for (int i = 1; i < str.size(); i++)
			if (str[i].second - str[i - 1].second - 1>maxlen)
				maxlen = str[i].second - str[i - 1].second - 1;
		return maxlen;
	}
};

accepted