87. Scramble String

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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.

判断两个字符串是不是scramble字符串，以某一点为中心对换位置，长度任意。abcd——bacd——acbd——cdab

递归判断从0开始到最后每个子串是不是，0-i内部是，或0-i与len-i是

public class Solution {    public boolean isScramble(String s1, String s2) {        if (s1.equals(s2)) return true;                 int[] letters = new int[26];        for (int i=0; i<s1.length(); i++) {            letters[s1.charAt(i)-'a']++;            letters[s2.charAt(i)-'a']--;        }        for (int i=0; i<26; i++) if (letters[i]!=0) 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;    }}

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