Wildcard Matching

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).

Example

isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

分析：

这题也是DP问题，横轴是S, 纵轴是P（含有?和*），那么我们可以得到：

if (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == ‘?‘) {
　　match[i][j] = match[i - 1][j - 1];
} else if (p.charAt(i - 1) == ‘*‘) {
　　match[i][j] = match[i - 1][j - 1] || match[i - 1][j] || match[i][j - 1];

　　// match[i][j - 1] 指的是用* 替换S中1个j或多个j之前的character，当然那些character可以是连续的。
}

 1 public class Solution {
 2     /**
 3      * @param s: A string
 4      * @param p: A string includes "?" and "*"
 5      * @return: A boolean
 6      */
 7     public boolean isMatch(String s, String p) {
 8         if (s == null || p == null)
 9             return false;
10         boolean[][] match = new boolean[p.length() + 1][s.length() + 1];
11         match[0][0] = true;
12         for (int i = 1; i < match[0].length; i++) {
13             match[0][i] = false;
14         }
15
16         for (int i = 1; i < match.length; i++) {
17             if (p.charAt(i - 1) == ‘*‘) {
18                 match[i][0] = match[i - 1][0];
19             }
20         }
21
22         for (int i = 1; i < match.length; i++) {
23             for (int j = 1; j < match[0].length; j++) {
24                 if (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == ‘?‘) {
25                     match[i][j] = match[i - 1][j - 1];
26                 } else if (p.charAt(i - 1) == ‘*‘) {
27                     match[i][j] = match[i - 1][j - 1] || match[i - 1][j] || match[i][j - 1];
28                 }
29             }
30         }
31
32         return match[p.length()][s.length()];
33     }
34 }