Longest Valid Parentheses
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DP:
从后向前遍历,若当前i是'(',则判断j=num[i]+1+i是否是')',若是则num[i]=num[i+1]+num[j+1]+2,若不是则不改变num[i]的值,仍是0,不能用num[i]=num[i+1]。
class Solution {public: int longestValidParentheses(string s) { int n=s.size(); if (!n) return 0; vector<int> num(n+1,0); int res=0; for (int i=n-2; i>=0; i--) { if (s[i]=='(') { int j=i+num[i+1]+1; if (j<n && s[j]==')') { num[i] = num[i+1] +2 + num[j+1]; res = res>num[i]?res:num[i]; } } } return res; }};
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