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.

 首先我想到的是scrambled string 就是原来string的全排列，而且似乎是这样的。所以算法如下：

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s1 == null) return false;
        HashMap<Integer,Integer> s1m = new HashMap<Integer,Integer>();
        for(int i = 0; i < s1.length(); i ++){
            Integer c = Integer.valueOf(s1.charAt(i));
            Integer cc= Integer.valueOf(s2.charAt(i));
            if(s1m.containsKey(c)){
                s1m.put(c, s1m.get(c) + 1);
            }else{
                s1m.put(c,1);
            }
            if(s1m.containsKey(cc)){
                s1m.put(cc,s1m.get(cc) - 1);
            }else{
                s1m.put(cc,-1);
            }
        }
        Iterator it= s1m.keySet().iterator();
        while (it.hasNext())
        {
            Object key=it.next();
            if(s1m.get(key) != 0){
                return false;
            }
        }
        return true;
    }
}

但是 测试是发现问题： 

input	output	expected
"abcd", "bdac"	true	false	X

看出来 并不是全排列，而是有相对顺序的。

 然后就想应该可以用DP来做，从顶往下走，一定在某一部能发现 s1 的左子树 等于 s2的右子树， s1的右子树 等于 s2的左子树。

 这是使用递归的算法：

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s1.equals(s2)) return true;
        int v1 = 0;
        int v2 = 0;
        int size = s1.length();
        for(int i = 0; i < size; i ++){
            v1 += Integer.valueOf(s1.charAt(i));
            v2 += Integer.valueOf(s2.charAt(i));
        }
        if(v1 != v2) return false;
        for (int i=1; i<size;i++) {
            if (isScramble(s1.substring(0,i), s2.substring(0,i)) && isScramble(s1.
substring(i), s2.substring(i)))
                return true;
            if (isScramble(s1.substring(0,i), s2.substring(size-i)) && isScramble(
s1.substring(i), s2.substring(0,size-i)))
                return true;
        }
        return false;
    }
}