【线性 dp】A016_LC_通配符匹配（分类讨论）

一、Problem

Given an input string (s) and a pattern §, implement wildcard pattern matching with support for ‘?’ and ‘*’.

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

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

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.

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

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

二、Solution

方法一：dp

经典串匹配，只不过加了几个约束条件…

• 定义状态：
  • $f[i][j]$ 表示 s 的前 $i$ 个字符能否用 p 的前 $j$ 个字符进行匹配，这样比较好理解
• 思考初始化：
  • $f[0][0] = true$
  • $f[0][j] = (f[0][j-1]\ and\ p[j-1] == *)$，当 s 为空串时，p 的前 $j$ 个字符只有是 * 号时，才能用 p 进行匹配；? 不行是因为 ? 只能匹配单个字符
• 思考状态转移方程：
  • 如果 $s[i] = p[j]$ 或者 $s[i] \not= p[j]$ 但 $p[j] =\ ?$，则有 $f[i][j] = f[i-1][j-1]$
  • 如果 $s[i] \not= p[j]，p[j] = *$ ，则 * 有两种作用：
    • 当做空串进行匹配（ac, ac*），$f[i][j] = f[i][j-1]$
    • 当做有意义的字符串（acac, ac*)，由上可得，因为 $f[3][3] = true$，所以 $f[i][j] = f[i-1][j]$
    • 综上，$f[i][j] = f[i][j-1]\ ||\ f[i-1][j]$
• 思考输出：$f[n][m]$

class Solution {
public:
    bool isMatch(string s, string p) {
    	int n = s.size(), m = p.size();
        vector<vector<bool>> f(n+1, vector<bool>(m+1));  f[0][0] = true;
        for (int j = 1; j <= m; j++) if (p[j-1] == '*') {
        	f[0][j] = f[0][j-1];
		}
		
        for (int i = 1; i <= n; i++)
    	for (int j = 1; j <= m; j++) {
    		if (s[i-1] == p[j-1] || p[j-1] == '?') {
    			f[i][j] = f[i-1][j-1];
    		} else if (p[j-1] == '*') {
    			f[i][j] = f[i][j-1] || f[i-1][j];
    		}
    	}
    	return f[n][m];
    }
};

复杂度分析

• 时间复杂度：$O(n × m)$，
• 空间复杂度：$O(n × m)$，