leetcode44

 class Solution {
     public boolean isMatch(String s, String p) {
         if (p == null || p.length() == 0) {
             return s == null || s.length() == 0;
         }
         boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
         dp[0][0] = true;
         int i, j;

         for (i = 1; i <= p.length(); i++) {
             if (p.charAt(i - 1) == '*') {
                 dp[0][i] = true;
             } else {
                 break;
             }
         }

         for (i = 1; i <= s.length(); i++) {
             for (j = 1; j <= p.length(); j++) {
                 // the letter of p[j] exactly match with s[i] or is '?'
                 if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') {
                     dp[i][j] = dp[i - 1][j - 1];
                 }

                 // the letter of p is '*'
                 if (p.charAt(j - 1) == '*') {
                     /*
                      * dp[i][j] = dp[i][j-1] means '*' act as empty card dp[i][j] = dp[i-1][j] means
                      * '*' act as wild card to match with s[i]
                      */
                     dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                 }
             }
         }
         return dp[s.length()][p.length()];
     }
 }

参考：https://leetcode.com/problems/wildcard-matching/discuss/429337/Java%3A-Wild-card!

补充一个python的实现：

 class Solution:
     def isMatch(self, s: str, p: str) -> bool:
         m,n = len(s),len(p)
         dp = [[False for _ in range(n+1)]for _ in range(m+1)]
         dp[0][0] = True

         for j in range(1,n+1):
             if p[j-1] == '*':
                 dp[0][j] = True
             else:
                 break
         for i in range(1,m+1):
             for j in range(1,n+1):
                 if s[i-1] == p[j-1] or p[j-1] == '?':
                     dp[i][j] = dp[i-1][j-1]
                 elif p[j-1] == '*':
                     dp[i][j] = dp[i-1][j] or dp[i][j-1]
         # print(dp)
         return dp[m][n]

思路：动态规划。

举例分析：s='acdcb' p='a*c?b' 。s长度为m，p长度为n。

初始化二维数组dp，有m+1行，n+1列。

dp[0][0] = True，先设置第0行，如果p中对应字符是‘*’，则设置为True，遇到非'*'后面都设置为False。

从第1行开始，如果s与p中对应字符是相同或者p中是'?'，则dp取其'左上角'元素的值。

如果p中的字符是'*'，则dp判断其'上方'和'左侧'的逻辑或，(即上方或左侧是否为True)。