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44.wildcard-matching

Statement

Metadata
  • Link: 通配符匹配
  • Difficulty: Hard
  • Tag: 贪心 递归 字符串 动态规划

给定一个字符串 (s) 和一个字符模式 (p) ，实现一个支持 '?' 和 '*' 的通配符匹配。

'?' 可以匹配任何单个字符。
'*' 可以匹配任意字符串（包括空字符串）。

两个字符串完全匹配才算匹配成功。

说明:

  • s 可能为空，且只包含从 a-z 的小写字母。
  • p 可能为空，且只包含从 a-z 的小写字母，以及字符 ? 和 *。

示例 1:

输入:
s = "aa"
p = "a"
输出: false
解释: "a" 无法匹配 "aa" 整个字符串。

示例 2:

输入:
s = "aa"
p = "*"
输出: true
解释: '*' 可以匹配任意字符串。

示例 3:

输入:
s = "cb"
p = "?a"
输出: false
解释: '?' 可以匹配 'c', 但第二个 'a' 无法匹配 'b'。

示例 4:

输入:
s = "adceb"
p = "*a*b"
输出: true
解释: 第一个 '*' 可以匹配空字符串, 第二个 '*' 可以匹配字符串 "dce".

示例 5:

输入:
s = "acdcb"
p = "a*c?b"
输出: false

Metadata
  • Link: Wildcard Matching
  • Difficulty: Hard
  • Tag: Greedy Recursion String Dynamic Programming

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*' where:

  • '?' Matches any single character.
  • '*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

 

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input: s = "cb", p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

 

Constraints:

  • 0 <= s.length, p.length <= 2000
  • s contains only lowercase English letters.
  • p contains only lowercase English letters, '?' 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:
    bool isMatch(string s, string p) {
        int n = s.length();
        int m = p.length();
        auto f = vector<vector<int>>(n + 1, vector<int>(m + 1, 0));
        f[0][0] = 1;
        for (int j = 1; j <= m; j++) {
            if (p[j - 1] != '*') {
                break;
            }

            f[0][j] = 1;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                char c_s = s[i - 1];
                char c_p = p[j - 1];
                if (c_p == '?' || c_p == '*' || c_s == c_p) {
                    f[i][j] |= f[i - 1][j - 1];
                }

                if (c_p == '*') {
                    f[i][j] |= f[i - 1][j] | f[i][j - 1];
                }
            }
        }
        return f[n][m];
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
    p = p.replace(/\?/g, ".");
    p = p.replace(/\*+/g, ".*");
    const r = new RegExp(`^${p}$`);
    return r.test(s);
};
最后更新: October 11, 2023
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