leetcode——Scramble_String

题目：

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.

思路：

题目中的树形结构也许会成为一个思路误区，使你直接建立树形节点去解题。其实，它仅仅是阐释一个递归方法。

但是题目中依然有个线索，及left1节点中字符的个数设为n1，right1节点中字符的个数设为n2；而另一个节点的left2字符个数为n3，right2字符个数为n4

那么，n1在不等于n2的情况下，不是等于n3就是n4.

因此，在递归的比较长度的left和right的子串的时候，截取的时候需要遵循以下规则：

for(int i=1;i<len;i++){            if((isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i))) ||                    (isScramble(s1.substring(0,i),s2.substring(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i)))){                return true;            }        }

而每个isScramble在判断是否相同或者可以互相转化的时候，首先判断长度是否相同；其次比较是否相等，相等直接返回true；

否则，继续判断是否所有字母个数都相同。如果对应字母个数相同，则调用上面的循环进行下一层的比较。

AC代码：

public boolean isScramble(String s1, String s2) {        int len = s1.length();        if(len!=s2.length())            return false;        if(s1.equals(s2))            return true;        int[] bools = new int[27];        for(char a:s1.toCharArray()){            bools[a-'a']++;        }        for(char b:s2.toCharArray()){            bools[b-'a']--;        }        for(Integer aa:bools)            if(aa!=0)                return false;        for(int i=1;i<len;i++){            if((isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i))) ||                    (isScramble(s1.substring(0,i),s2.substring(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i)))){                return true;            }        }        return false;    }