LongestValidParentheses.h

/*
 Author:     Annie Kim, anniekim.pku@gmail.com
 Date:       May 6, 2013
 Update:     Sep 26, 2013
 Problem:    Longest Valid Parentheses
 Difficulty: Easy
 Source:     http://leetcode.com/onlinejudge#question_32
 Notes:
 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.

 Solution: O(n).
 */

class Solution {
public:
    int longestValidParentheses(string s) {
        return longestValidParentheses_1(s);
    }
    
    // Solution 1, from fuwutu. Smart!
    // push the length of last valid parentheses into stack.
    int longestValidParentheses_1(string s) {
        stack<int> stk;
        int res = 0, count = 0;
        for(int i = 0; i < s.size(); ++i) {
            if (s[i] == '(') {
                stk.push(count);
                count = 0;
            } else if (!stk.empty()) {
                count += (1 + stk.top());
                stk.pop();
                res = max(res, count);
            } else {
                count = 0;
            }
        }
        return res * 2;
    }
    
    // Solution 2, my original version.
    // Traverse the string twice, taking O(n) time.
    // First time, mark all the matching parentheses as '*' (push the index of '(' into <stack>).
    // Second time, count the longest consecutive '*'.
    int longestValidParentheses_2(string s) {
        stack<int> stk;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '(') {
                stk.push(i);
            } else if (!stk.empty()) {
                s[stk.top()] = '*';
                s[i] = '*';
                stk.pop();
            }
        }
        int res = 0, part = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == '*') {
                part++;
            } else {
                res = max(res, part);
                part = 0;
            }
        }
        res = max(res, part);
        return res;
    }
};