LeetCode 65 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

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.

分析：

网上有动规的解法，看了解释也能看明白，只是 我觉得自己看到这题，没有办法想到用动规做，所以我还是贴出自己的递归代码。

既然题目都把字符串表示成二叉树的形式，我当然首先想到的是递归了。

把s1分成s11和s12，s2分成s21和s22，则有

isScramble(s11, s21) && isScramble(s12, s22)

或者

isScramble(s11, s22) && isScramble(s12, s21)

怎么分割，我们不知道，所以要试验每一个位置，只要有一个位置的分割可行，就可以返回true.

对于递归解法来讲，要剪枝才能过，剪枝的意思就是把不合理的尽早去除，不要再进入递归了。

比如：

1，长度不相等，

2，排序之后内容不相同

还有，一种能尽早得到结果，比如

s1.equals(s2)

这也没有必要进入下层递归了。

另外，我们知道，在适用的情况下，桶排是最快的排序。

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length()) return false;
        if(s1.equals(s2)) return true;
        int[] count = new int[26];
        int len = s1.length();
        for(int i=0; i<len; i++){
            count[s1.charAt(i)-'a']++;
            count[s2.charAt(i)-'a']--;
        }
        for(int i=0; i<26; i++){
            if(count[i]!=0) return false;
        }
    
        for(int step=1; step<len; step++){
            String s11 = s1.substring(0,step);
            String s12 = s1.substring(step);
            
            String s21 = s2.substring(0,step);
            String s22 = s2.substring(step);
            if(isScramble(s11,s21)&&isScramble(s12,s22)) return true;
            
            s21 = s2.substring(0, len-step);
            s22 = s2.substring(len-step);
            if(isScramble(s11,s22)&&isScramble(s12,s21)) return true;
        }
        return false;
    }
}