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 }