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32.longest-valid-parentheses

Statement

Metadata

给你一个只包含 '(' 和 ')' 的字符串,找出最长有效(格式正确且连续)括号子串的长度。

 

示例 1:

输入:s = "(()"
输出:2
解释:最长有效括号子串是 "()"

示例 2:

输入:s = ")()())"
输出:4
解释:最长有效括号子串是 "()()"

示例 3:

输入:s = ""
输出:0

 

提示:

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

Metadata

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
// head

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

最后更新: January 15, 2023
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