LeetCode 87.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.

public class Solution {    public boolean isScramble(String s1, String s2) {        char[] t1 = s1.toCharArray();        char[] t2 = s2.toCharArray();        if (t1.length != t2.length) return false;        return scramble(t1, 0, t2, 0, t2.length);    }    //此方法做出来一下这几点的优化方案：    //用 字符数组和长度来替代字符串不用每次都切割子串    // 在判断字符的数组中的元素是否相同的时候少了一个对letter的循环遍历    private  boolean scramble(char[] s1, int i, char[] s2, int j, int n) {        //当 n == 1 的时候应该向上回溯        if (n == 1) return s1[i] == s2[j];        //判断字符是否相等        int[] letter = new int[26];        for (int k = i; k < i+n; k++) {            ++letter[s1[k]-'a'];        }        for (int k = j; k < j+n; k++) {            //存在不想等的字符            if (--letter[s2[k]-'a'] < 0)                return false;        }        for (int l = 1; l < n; l++) {            //将字符数组的前半部分和后半部分分别的进行对比 相当于s.substring()            if (scramble(s1,i,s2,j,l)                    && scramble(s1,i+l,s2,j+l,n-l)) return true;            //将s1前半部分和s2的后半部分对比，            if (scramble(s1,i,s2,n+j-l,l)                    && scramble(s1,i+l,s2,j,n-l)) return true;        }        return false;    }}