【Leetcode】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.

思路：使用递归的方法，判断s1和s2的子串是否分别可以进行重组，如果都可以，则返回true。否则，返回false。

代码：

class Solution {public:    bool isScramble(string s1, string s2) {        string s1_temp = s1, s2_temp = s2;        if(s1==s2)            return true;        if(s1.length()!=s2.length())            return false;        sort(s1_temp.begin(),s1_temp.end());        sort(s2_temp.begin(),s2_temp.end());        if(s1_temp!=s2_temp)            return false;        int size = s1.length();        for(int i=1;i<size;i++)        {            string s1left = s1.substr(0,i);            string s2left = s2.substr(0,i);            string s1right = s1.substr(i);            string s2right = s2.substr(i);            if(isScramble(s1left,s2left)&&isScramble(s1right,s2right))                return true;            s2left = s2.substr(0, size-i);            s2right = s2.substr(size-i);            if(isScramble(s1left,s2right)&&isScramble(s1right,s2left))                return true;        }        return false;    }};