[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:

 public class Solution {
     public boolean isScramble(String s1, String s2) {
         if(s1.equals(s2))
             return true;
         int l1=s1.length();
         int l2=s2.length();
         if(l1!=l2)
             return false;
         if(l1==0)
             return true;
         if(l1==1)
             return s1.equals(s2);
         char[] c_s1=s1.toCharArray();
         char[] c_s2=s2.toCharArray();
         Arrays.sort(c_s1);
         Arrays.sort(c_s2);
         for(int i=0;i<c_s1.length;++i){
             if(c_s1[i]!=c_s2[i])
                 return false;
         }
         boolean b=false;
         for(int i=1;i<l1&&!b;++i){
             String s11=s1.substring(0,i);
             String s12=s1.substring(i);
             String s21=s2.substring(0,i);
             String s22=s2.substring(i);
             b=isScramble(s11, s21)&&isScramble(s12, s22);
             if(!b){
                 String s31=s2.substring(0, l1-i);
                 String s32=s2.substring(l1-i);
                 b=isScramble(s11, s32)&&isScramble(s12, s31);
             }
         }
         return b;
     }
 }

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20150220:

 public class Solution {
     public boolean isScramble(String s1, String s2) {
         if(s1==null||s2==null||s1.length()!=s2.length())
             return false;
         if(s1.equals(s2))
             return true;
         char[] c1=s1.toCharArray();
         char[] c2=s2.toCharArray();
         Arrays.sort(c1);
         Arrays.sort(c2);
         for(int i=0;i<c1.length;++i){
             if(c1[i]!=c2[i])
                 return false;
         }
         int N=s1.length();
         for(int i=1;i<N;++i){
             String s1_temp1=s1.substring(0,i);
             String s1_temp2=s1.substring(i);
             String s2_temp1=s2.substring(0,i);
             String s2_temp2=s2.substring(i);
             if(isScramble(s1_temp1, s2_temp1)&&isScramble(s1_temp2, s2_temp2))
                 return true;
             String s2_temp3=s2.substring(0,N-i);
             String s2_temp4=s2.substring(N-i);
             if(isScramble(s1_temp1, s2_temp4)&&isScramble(s1_temp2, s2_temp3))
                 return true;
         }
         return false;
     }
 }