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.

Scramble String
 1 public class Solution {
 2         public boolean isScramble(String s1, String s2) {
 3             int len1 = s1.length();
 4             int len2 = s2.length();
 5             if(len1!=len2) return false;
 6             char[] c = new char[26];
 7             for(int i=0;i<len1;i++){
 8                 c[s1.charAt(i)-‘a‘]++;
 9             }
10             for(int i=0;i<len2;i++){
11                 c[s2.charAt(i)-‘a‘]--;
12             }
13             for(int i=0;i<26;i++){
14                 if(c[i]!=0) return false;
15             }
16             if(len1==len2 &&len1==1) return true;
17             for(int i=0;i<len1;i++){
18                 boolean res= isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i));
19                         res |=isScramble(s1.substring(0,i),s2.substring(len2-i))&&isScramble(s1.substring(i),s2.substring(0,len2-i));
20                 if(res) return true;
21             }
22             return false;
23             
24         }
25     }
View Code

Scramble String

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