Longest Valid ParenthesesMar 1 '125700 / 20657
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.
之前大摩面试时,问到了这题,当时愚昧了。
一开始,写错了。
后来,想了个N^2的方法。 T_T。
不过好歹这次一次AC :)
struct A { A(char cc, int vv=0) : c(cc), v(vv) {} char c; int v; }; class Solution { public: int longestValidParentheses(string s) { int len = s.size(); if (len == 0) return 0; stack<A> st; st.push(A(s[0])); int cnt = 0; for (int i = 1; i < len; i++) { cnt = 0; if (s[i] == '(') st.push(A(s[i])); else { while (!st.empty() && st.top().c == '#') { cnt += st.top().v; st.pop(); } if (!st.empty() && st.top().c == '(') { cnt += 2; st.pop(); st.push(A('#', cnt)); } else { // top().c == '#' if (cnt > 0) st.push(A('#',cnt)); st.push(A(')')); } } } cnt = 0; int mx = 0; while (!st.empty()) { if (st.top().c == '#') { cnt += st.top().v; if (mx < cnt) mx = cnt; } else { cnt = 0; } st.pop(); } return mx; } };
class Solution { public: int longestValidParentheses(string s) { int n = s.size(); if (n == 0) return 0; int i = 0; stack<pair<char, int> > st; st.push(make_pair('#', 0)); st.push(make_pair(s[i++], 0)); while (!st.empty() && i < n) { if (s[i] == '(') st.push(make_pair(s[i++], 0)); else { i++; int cnt = 0; while (st.top().first == '.') cnt += st.top().second, st.pop(); if (st.top().first == '(') { st.pop(); st.push(make_pair('.', cnt + 1)); } else { if (cnt > 0) st.push(make_pair('.', cnt)); st.push(make_pair(')', 0)); } } } int res = 0; while (st.top().first != '#') { int cnt = 0; while (st.top().first == '.') cnt += st.top().second, st.pop(); res = max(res, cnt); while (st.top().first == '(' || st.top().first == ')') st.pop(); } return res*2; } };