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.
解题思路:这道题有多种方法可以做,最后采用了动态规划。但普通的动态规划都是向前考虑的,这道题是向后考虑的,可以说是逆向的。比如普通动态规划dp[i]应该是从位置0~i的最长有效括号长度,而这道题的dp[i]表示的是从i到length-1的最长有效括号长度。
因此i从后往前推,若s[i]=’(‘,则匹配的右括号位置end应该是i + dp[i + 1] + 1,若该位置确实是右括号则dp[i] = dp[i + 1] + 2,另外为了计算连续的有效括号长度需要加上dp[end + 1]。
代码:
class Solution {
public:
int longestValidParentheses(string s) {
int len = s.length();
int *dp = new int[len];
memset(dp,0,len*sizeof(int));
for (int i = len - 2; i >= 0; i--) {
if (s[i] == '(') {
int end = i + dp[i + 1] + 1;
if (end<len && s[end] == ')') {
dp[i] = dp[i + 1] + 2;
if (end<len - 1) {
dp[i] += dp[end + 1];
}
}
}
}
int maxLen = 0;
for (int i = 0; i<len; i++) {
if (dp[i]>maxLen)
maxLen = dp[i];
}
return sizeof(dp);
delete[]dp;
}
};