# 32.longest-valid-parentheses

## Statement

• Difficulty: Hard
• Tag: `栈` `字符串` `动态规划`

``````输入：s = "(()"

``````

``````输入：s = ")()())"

``````

``````输入：s = ""

``````

• `0 <= s.length <= 3 * 104`
• `s[i]``'('``')'`

• Difficulty: Hard
• Tag: `Stack` `String` `Dynamic Programming`

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 * 104`
• `s[i]` is `'('`, or `')'`.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
int longestValidParentheses(string s) {
int n = s.length();

auto v = vector<int>();
auto f = vector<int>(n + 1, 0);

for (int i = 0; i < s.length(); i++) {
if (s[i] == '(') {
v.push_back(i);
} else {
if (!v.empty()) {
f[i + 1] = i - v.back() + 1;
v.pop_back();
}
}
}

int res = 0;
for (int i = 1; i <= n; i++) {
f[i] += f[i - f[i]];
res = max(res, f[i]);
}

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````