[leetcode]Longest Valid Parentheses

32. 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.

DP solution:

// dp[i] represents the longest valid parentheses ending at 
// i-1 th character: s[i - 1]
int longestValidParentheses(string s)
{
    int N = s.length();
    if (N <= 1) return 0;
    vector<int> dp(N + 1, 0);
    for (int i = 2; i <= N; ++i)
    {
        // if s[i - 1] == '(', there will be no valid
        // parentheses ending at s[i - 1]
        if (s[i - 1] == '(')
            dp[i] = 0;
        else
        {
            // if there exists valid parentheses ending at s[i - 2] and the
            // character before the valid parentheses is '('.
            // (()) i = 4, dp[3] = 2, s[0] == '(', then dp[4] = dp[3] + 2 = 4
            if (dp[i - 1] && i - dp[i - 1] > 1 && s[i - 2 - dp[i - 1]] == '(')
                dp[i] = 2 + dp[i - 1];
            // no valid ending at s[i - 2] but s[i - 2] == '('
            // (), i = 2, dp[i] = 2 
            else if (dp[i - 1] == 0 && s[i - 2] == '(')
                dp[i] = 2;
            else
                dp[i] = 0;
        }
        // add preceding valid parentheses.
        // ()(()) i = 6, dp[6] = 2 + dp[5] + dp[2] = 6
        // ()() i = 4, dp[4] = 2 + dp[2]
        if (dp[i] && i - 1 - dp[i] > 0 && dp[i - dp[i]])
            dp[i] += dp[i - dp[i]];
    }
    int res = 0;
    for (auto len : dp)
        res = max(len, res);
    return res;
}

stack solution
在栈中存没配对的括号的位置，则夹在相邻两个未配对括号中的是配对的。

int longestValidParentheses(string s)
{
    int N = s.length();
    if (N <= 1) return 0;
    stack<int> st;
    int res = 0;
    for (int i = 0; i < N; ++i)
        if (s[i] == '(')
            st.push(i);
        else if (st.size() && s[st.top()] == '(')
            st.pop();
        else
            st.push(i);
        if (st.empty())
            res = N;
        else
        {
            int a = N, b = 0;
            while (st.size())
            {
                b = st.top();
                st.pop();
                res = max(res, a - b - 1);
                a = b;
            }
            res = max(res, a);
        }
        return res;
}