LeetCode OJ - Interleaving String

这道题用DP来解决，也可以用递归，但是时间复杂度很高，被报TLE。

下面是DP的AC代码。

 1 /**
 2      * Solution 2 :DP Accepted
 3      * create a (len1+1) x (len2+1) matrix A, A[i][j] means s3[0...i+j-1] is formed by the interleaving of s1[0...i-1] and s2[0...j-1]
 4      * A[i][j] =  True   if A[i-1][j] = true && s3[i+j-1] = s1[i-1] or A[i][j-1] = true && s3[i+j-1] = s2[j-1];
 5      *            False  else
 6      * @param s1
 7      * @param s2
 8      * @param s3
 9      * @return
10      */
11     public boolean isInterleave2(String s1, String s2, String s3){
12         int len1 = s1.length();
13         int len2 = s2.length();
14         int len3 = s3.length();
15         
16         if(len1+len2!=len3)
17             return false;
18         
19         char[] c1 = s1.toCharArray();
20         char[] c2 = s2.toCharArray();
21         char[] c3 = s3.toCharArray();
22         
23         //DP
24         boolean[][] A = new boolean[len1+1][len2+1];
25         A[0][0] = true;
26         for(int i = 1;i<=len1;i++)
27             if( A[i-1][0]== true && c3[i-1] == c1[i-1])
28                 A[i][0] = true;
29             else
30                 A[i][0] = false;
31         for(int j=1;j<=len2;j++)
32             if(A[0][j-1] == true && c3[j-1] == c2[j-1])
33                 A[0][j] = true;
34             else
35                 A[0][j] = false;
36         
37         for(int i=1;i<=len1;i++){
38             for(int j=1;j<=len2;j++){
39                 if(A[i-1][j] == false && A[i][j-1] == false)
40                     A[i][j] = false;
41                 else if(A[i-1][j] == true && c3[i+j-1] == c1[i-1])
42                     A[i][j] = true;
43                 else if(A[i][j-1] == true && c3[i+j-1] == c2[j-1])
44                     A[i][j] = true;
45                 else
46                     A[i][j] = false;
47             }
48         }
49         return A[len1][len2];
50     }

 

posted @ 2014-05-06 12:15  echoht  阅读(129)  评论(0)  编辑  收藏  举报