LeetCode: Wildcard Matching 解题报告

 1 public boolean isMatch1(String s, String p) {
 2         if (s == null || p == null) {
 3             return false;
 4         }
 5         
 6         int lens = s.length();
 7         int lenp = p.length();
 8         
 9         // 创建一个Dp二维数组
10         boolean[][] D = new boolean[lens + 1][lenp + 1];
11         
12         boolean flag = false;
13         
14         for (int i = 0; i <= lens; i++) {
15             flag = false;
16             for (int j = 0; j <= lenp; j++) {
17                 // both is empty.
18                 if (i == 0 && j == 0) {
19                     D[i][j] = true;
20                     flag = true;
21                     continue;
22                 }
23                 
24                 // if P is empty, s is not empty, it is false.
25                 if (j == 0) {
26                     D[i][j] = false;
27                     continue;
28                 }
29                 
30                 // if S is empty, P is not empty
31                 if (i == 0) {
32                     D[i][j] = D[i][j - 1] && p.charAt(j - 1) == ‘*‘;
33                 } else {
34                     D[i][j] = (matchChar(s.charAt(i - 1), p.charAt(j - 1)) && D[i - 1][j - 1])
35                       || (p.charAt(j - 1) == ‘*‘ && (D[i][j - 1] || D[i - 1][j]));    
36                 }
37                 
38                 if (D[i][j]) {
39                     flag = true;
40                 }
41                 
42                 // Greedy. 在此即可以退出，因为* 可以匹配余下的所有的字符串。    
43                 if (D[i][j] && p.charAt(j - 1) == ‘*‘ && j == lenp) {
44                     return true;
45                 }
46             }
47             
48             if (!flag) {
49                 return false;
50             }
51         }
52         
53         return D[lens][lenp];
54     }
55     
56     public static boolean matchChar(char c, char p) {
57         return (p == ‘?‘ || p == c);
58     }
59 }

 1     public static boolean matchChar(char c, char p) {
 2         return (p == ‘?‘ || p == c);
 3     }
 4     
 5     public static boolean isMatch(String s, String p) {
 6         if (s == null || p == null) {
 7             return false;
 8         }
 9         
10         int indexS = 0;
11         int indexP = 0;
12         
13         int lenS = s.length();
14         int lenP = p.length();
15         
16         int preS = 0;
17         int preP = 0;
18         
19         boolean back = false;
20         
21         while (indexS < lenS) {
22             if (indexP < lenP && matchChar(s.charAt(indexS), p.charAt(indexP))) {
23                 indexS++;
24                 indexP++;
25             } else if (indexP < lenP && p.charAt(indexP) == ‘*‘) {
26                 while (indexP < lenP && p.charAt(indexP) == ‘*‘) {
27                     indexP++;
28                 }
29                 
30                 //P的最后一个是 *，表示可以匹配任何字符串
31                 if (indexP == lenP) {
32                     return true;
33                 }
34                 
35                 // 记录下这个匹配位置。
36                 back = true;
37                 preS = indexS;
38                 preP = indexP;
39             } else if (back) {
40                 indexS = ++preS;
41                 indexP = preP;
42             } else {
43                 return false;
44             }
45         }
46         
47         // 跳过末尾的所有的*.
48         while (indexP < lenP && p.charAt(indexP) == ‘*‘) {
49             indexP++;
50         }
51         
52         if (indexS == lenS && indexP == lenP) {
53             return true;
54         }
55 
56         return false;
57     }