[Leetcode] Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    
  gr    eat
 /     /  
g   r  e   at
           / 
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    
  rg    eat
 /     /  
r   g  e   at
           / 
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    
  rg    tae
 /     /  
r   g  ta  e
       / 
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Solution:

 1 public class Solution {
 2     public boolean isScramble(String s1, String s2) {
 3         if(s1.equals(s2))
 4             return true;
 5         int l1=s1.length();
 6         int l2=s2.length();
 7         if(l1!=l2)
 8             return false;
 9         if(l1==0)
10             return true;
11         if(l1==1)
12             return s1.equals(s2);
13         char[] c_s1=s1.toCharArray();
14         char[] c_s2=s2.toCharArray();
15         Arrays.sort(c_s1);
16         Arrays.sort(c_s2);
17         for(int i=0;i<c_s1.length;++i){
18             if(c_s1[i]!=c_s2[i])
19                 return false;
20         }
21         boolean b=false;
22         for(int i=1;i<l1&&!b;++i){
23             String s11=s1.substring(0,i);
24             String s12=s1.substring(i);
25             String s21=s2.substring(0,i);
26             String s22=s2.substring(i);
27             b=isScramble(s11, s21)&&isScramble(s12, s22);
28             if(!b){
29                 String s31=s2.substring(0, l1-i);
30                 String s32=s2.substring(l1-i);
31                 b=isScramble(s11, s32)&&isScramble(s12, s31);
32             }
33         }
34         return b;
35     }
36 }

------------------------------------------------------------------------------------

20150220:

 1 public class Solution {
 2     public boolean isScramble(String s1, String s2) {
 3         if(s1==null||s2==null||s1.length()!=s2.length())
 4             return false;
 5         if(s1.equals(s2))
 6             return true;
 7         char[] c1=s1.toCharArray();
 8         char[] c2=s2.toCharArray();
 9         Arrays.sort(c1);
10         Arrays.sort(c2);
11         for(int i=0;i<c1.length;++i){
12             if(c1[i]!=c2[i])
13                 return false;
14         }
15         int N=s1.length();
16         for(int i=1;i<N;++i){
17             String s1_temp1=s1.substring(0,i);
18             String s1_temp2=s1.substring(i);
19             String s2_temp1=s2.substring(0,i);
20             String s2_temp2=s2.substring(i);
21             if(isScramble(s1_temp1, s2_temp1)&&isScramble(s1_temp2, s2_temp2))
22                 return true;
23             String s2_temp3=s2.substring(0,N-i);
24             String s2_temp4=s2.substring(N-i);
25             if(isScramble(s1_temp1, s2_temp4)&&isScramble(s1_temp2, s2_temp3))
26                 return true;
27         }
28         return false;
29     }
30 }