20.valid-parentheses

Statement

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Metadata

• Link: 有效的括号
• Difficulty: Easy
• Tag: 栈 字符串

给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串 s ，判断字符串是否有效。

有效字符串需满足：

1. 左括号必须用相同类型的右括号闭合。
2. 左括号必须以正确的顺序闭合。

 

示例 1：

输入：s = "()"
输出：true

示例 2：

输入：s = "()[]{}"
输出：true

示例 3：

输入：s = "(]"
输出：false

示例 4：

输入：s = "([)]"
输出：false

示例 5：

输入：s = "{[]}"
输出：true

 

提示：

• 1 <= s.length <= 104
• s 仅由括号 '()[]{}' 组成

Metadata

• Link: Valid Parentheses
• Difficulty: Easy
• Tag: Stack String

Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

 

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "()[]{}"
Output: true

Example 3:

Input: s = "(]"
Output: false

 

Constraints:

• 1 <= s.length <= 104
• s consists of parentheses only '()[]{}'.

Solution

C++

#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:
    bool isValid(string s) {
        auto mp = map<char, char>({
                {'(', ')'},
                {'[', ']'},
                {'{', '}'},
        });

        auto v = vector<char>();

        for (const auto &c : s) {
            if (c == '(' || c == '[' || c == '{') {
                v.push_back(c);
            } else {
                if (v.empty()) {
                    return false;
                }

                if (mp[v.back()] != c) {
                    return false;
                }

                v.pop_back();
            }
        }

        return v.empty();
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

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最后更新: October 11, 2023