leetcode87: 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".

这道题目定义了字符串变换的某种规则，并要求判断这种规则下，这两个字符串是否相等。

变化的规则主要是不定时地交换左右两个子树。

所以我们可以通过递归的方式，从递归边界判断两个最小的子树是否相等，在逐步往上判断。

首先对字符串进行初步的判断，字符串是否相等，若字符串不相等，字符串的内容是否相等。

如果S1与S2的左右子树没有交换，则用第一个if的规则进行判断。

如果S1与S2的左右子树发生了交换，则用第二个if的规则进行判断。

解题代码如下：

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