Scramble String.java

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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
    	int length1 = s1.length();
	int length2 = s2.length();
	if (length1 != length2)
	    return false;

	if (length1 == 0 || s1.equals(s2)) return true;

	char[] ca1 = s1.toCharArray();
	char[] ca2 = s2.toCharArray();
	Arrays.sort(ca1);
	Arrays.sort(ca2);
	if (!Arrays.equals(ca1, ca2)) return false;

	int i = 1;
	while (i < length1) {
	    String a1 = s1.substring(0, i);
	    String b1 = s1.substring(i, length1);
	    String a2 = s2.substring(0, i);
            String b2 = s2.substring(i, length2);
	    if (a1.equals(b2) && b1.equals(a2)) return true;
	    boolean r = isScramble(a1, a2) && isScramble(b1, b2);
	    if (!r) {
		String c2 = s2.substring(0, length1 - i);
		String d2 = s2.substring(length1 - i);
		r = isScramble(a1, d2) && isScramble(b1, c2);
	    }
	    if (r) return true;
	    i++;
	}
	return false;
    }
}