Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
Example 1:
Input: s = "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()".
Example 2:
Input: s = ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()".
Example 3:
Input: s = ""
Output: 0
Constraints:
0 <= s.length <= 3 * 10^4
s[i] is '(', or ')'.
Solution 1: Dynamic Programming
class Solution {
public:
int longestValidParentheses(string s) {
int n = s.size();
int res = 0;
vector<int> dp(n, 0);
for (int i = 0; i < n; i++) {
if (i > 0 && s[i] == ')') {
if (s[i-1] == '(') {
int val = i - 2 >= 0 ? dp[i-2] : 0;
dp[i] = 2 + val;
} else {
int ind = i - dp[i-1] - 1;
if (ind >= 0 && s[ind] == '(') {
int val = ind > 0 ? dp[ind - 1] : 0;
dp[i] = 2 + dp[i-1] + val;
}
}
res = max(res, dp[i]);
}
}
return res;
}
};
Solution 2: Stack
class Solution {
public:
int longestValidParentheses(string s) {
stack<int> mystack;
mystack.push(-1);
int res = 0;
for (int i = 0; i < s.size(); i++) {
int t = mystack.top();
if (t != -1 && s[i] == ')' && s[t]=='(') {
mystack.pop();
res = max(res, i - mystack.top());
} else {
mystack.push(i);
}
}
return res;
}
};