[leetcode]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[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;
}

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);
<
998b
span class="hljs-keyword">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;
}