[leetcode]Scramble String

问题描述：

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 ifs2 is a scrambled string of s1.

基本思想：

递归的分别验证树的左子树和右子树。

代码：

   public boolean isScramble(String s1, String s2) {  //java        if(s1 == null || s2 == null || s1.length() != s2.length())            return false;                if(s1.equals(s2))            return true;        char [] chs1= s1.toCharArray();        char [] chs2= s2.toCharArray();        Arrays.sort(chs1);        Arrays.sort(chs2);        if(!Arrays.equals(chs1, chs2))        return false;                for(int i = 1; i < s1.length(); 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(s2.length()-i))&&isScramble(s1.substring(i),s2.substring(0,s2.length()-i)))                return true;        }        return false;    }