leetcode Longest Valid Parentheses
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Longest Valid Parentheses 原题地址:
https://oj.leetcode.com/problems/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.
这道题其实可以看成是求数组最大子数组之和的扩展。
解法见这里http://blog.csdn.net/cfc1243570631/article/details/9304525
假设输入括号表达式为String s,维护一个长度为s.length的一维数组dp[],数组元素初始化为0。 dp[i]表示从s[i]到s[s.length - 1]包含s[i]的最长的有效匹配括号子串长度。则存在如下关系:
- dp[s.length - 1] = 0;
- i从n - 2 -> 0逆向求dp[],并记录其最大值。若s[i] == '(',则在s中从i开始到s.length - 1计算dp[i]的值。这个计算分为两步,通过dp[i + 1]进行的(注意dp[i + 1]已经在上一步求解):
- 在s中寻找从i + 1开始的有效括号匹配子串长度,即dp[i + 1],跳过这段有效的括号子串,查看下一个字符,其下标为j = i + 1 + dp[i + 1]。若j没有越界,并且s[j] == ‘)’,则s[i ... j]为有效括号匹配,dp[i] =dp[i + 1] + 2。
- 在求得了s[i ... j]的有效匹配长度之后,若j + 1没有越界,则dp[i]的值还要加上从j + 1开始的最长有效匹配,即dp[j + 1]。
public class Solution { public int longestValidParentheses(String s) { int len = s.length();int[] dp = new int[len];int max = 0;for (int i = len - 2; i >= 0; i--) {if (s.charAt(i) == '(') {int j = i + 1 + dp[i+1];if (j < len && s.charAt(j) == ')') {dp[i] = dp[i+1] +2;if (j + 1 < len)dp[i] += dp[j+1];}}max = Math.max(max, dp[i]);}return max; }}
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