Interleaving String

 1 public class Solution {
 2     public boolean isInterleave(String s1, String s2, String s3) {
 3         // IMPORTANT: Please reset any member data you declared, as
 4         // the same Solution instance will be reused for each test case.
 5        int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
 6         if(len3 != len1 + len2){
 7             return false;
 8         }
 9         
10         boolean[][] match = new boolean[len1 + 1][len2 + 1];
11         match[0][0] = true;
12         
13         // j == 0
14         int i = 1;
15         while(i <= len1 && s1.charAt(i-1) == s3.charAt(i-1)){
16             match[i][0] = true;
17             i++;
18         }
19         
20         // i == 0
21         int j = 1;
22         while(j <= len2 && s2.charAt(j-1) == s3.charAt(j-1)){
23             match[0][j] = true;
24             j++;
25         }
26         
27         for(int m = 1; m <= len1; m++){
28             for(int n = 1; n <= len2; n++){
29                 char c = s3.charAt(m + n - 1);
30                 if(c == s1.charAt(m - 1)){
31                     match[m][n] |= match[m-1][n];
32                 }
33                 if(c == s2.charAt(n - 1)){
34                     match[m][n] |= match[m][n-1];
35                 }
36             }
37         }
38         return match[len1][len2];
39     }
40 }

 1 public class Solution {
 2     public boolean isInterleave(String s1, String s2, String s3) {
 3         // IMPORTANT: Please reset any member data you declared, as
 4         // the same Solution instance will be reused for each test case.
 5        int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
 6         if(len3 != len1 + len2){
 7             return false;
 8         }
 9         
10         boolean[][] match = new boolean[len1 + 1][len2 + 1];
11         match[0][0] = true;
12         
13         // j == 0
14         int i = 1;
15         while(i <= len1 && s1.charAt(i-1) == s3.charAt(i-1)){
16             match[i][0] = true;
17             i++;
18         }
19         
20         // i == 0
21         int j = 1;
22         while(j <= len2 && s2.charAt(j-1) == s3.charAt(j-1)){
23             match[0][j] = true;
24             j++;
25         }
26         
27         for(int m = 1; m <= len1; m++){
28             for(int n = 1; n <= len2; n++){
29                 char c = s3.charAt(m + n - 1);
30                 if(c == s1.charAt(m - 1)){
31                     match[m][n] |= match[m-1][n];
32                 }
33                 if(c == s2.charAt(n - 1)){
34                     match[m][n] |= match[m][n-1];
35                 }
36             }
37         }
38         return match[len1][len2];
39     }
40 }

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原文地址：https://www.cnblogs.com/jasonC/p/3433319.html