[LeetCode 44.] 通配符匹配

LeetCode 44. 通配符匹配

一道简化版的正则表达式匹配题。

题目描述

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'.

Example 4:

Input: s = "adceb", p = "ab"
Output: true
Explanation: The first '' matches the empty sequence, while the second '' matches the substring "dce".

Example 5:

Input: s = "acdcb", p = "a*c?b"
Output: false

Constraints:

• 0 <= s.length, p.length <= 2000
• s contains only lowercase English letters.
• p contains only lowercase English letters, '?' or '*'.

解题思路

这道题是对 LeetCode 10. 正则表达式匹配 的简化，区别在于：

• 使用 '?' 代替正则表达式中的 '.'
• 使用单个字符 '' 来匹配0个或任意多个字符，正则表达式是用 "a" / ".*" 两个字符才能匹配0个或多个字符。

这就简化了题目，遇到 '' 的时候不用向前看一个字符了，可以直接原地匹配0个字符、1个字符、多个字符。本题的考点就在于 '' 匹配2个及以上字符的状态转移方程。

先做这道题，在做正则表达式匹配的题，可能会更顺利一些。

参考代码

/*
 * @lc app=leetcode id=44 lang=cpp
 *
 * [44] Wildcard Matching
 */

// @lc code=start
class Solution {
public:
    bool isMatch(string s, string p) {
        if (!s.empty() && p.empty()) return false;

        size_t ns = s.size(), np = p.size();
        vector<vector<bool>> dp(ns+1, vector<bool>(np+1));
        dp[0][0] = true;
        for (int i=1; i<=ns; i++) {
            dp[i][0] =false;
        }
        for (int j=1; j<=np; j++) {
            dp[0][j] = dp[0][j-1] && (p[j-1] == '*');
        }

        for (int i=1; i<=ns; i++) {
            for (int j=1; j<=np; j++) {
                if (s[i-1] == p[j-1] || p[j-1] == '?') {
                    dp[i][j] = dp[i-1][j-1];
                } else if (p[j-1] == '*') {
                    // match 0, 1, or n char
                    dp[i][j] = dp[i][j-1] || dp[i-1][j-1] || dp[i-1][j];
                } else {
                    dp[i][j] = false;
                }
            }
        }
        return dp[ns][np];
    } // AC
};
// @lc code=end