[leetcode] Longest Valid Parentheses
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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.
参考链接:http://blog.csdn.net/abcbc/article/details/8826782
思路:
这道题可以用一维动态规划逆向求解。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]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;
- 从i - 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s[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]。
代码:
class Solution {public: int longestValidParentheses(string s) { int n=s.size(); int dp[n]; memset(dp,0,sizeof(dp)); int res=0; for(int i=n-2;i>=0;i--){ if(s[i]=='('){ int j=i+1+dp[i+1]; if(j<n && s[j]==')'){ dp[i]=dp[i+1]+2; int k=0; if(j+1<n){ k=dp[j+1]; } dp[i]+=k; } res=max(res,dp[i]); } } return res; }};
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