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.

程序：

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1 == s2)            return true;        int c[26] = {0};        for(int i=0;i<s1.length();i++)        {            c[s1[i]-'a']++;            c[s2[i]-'a']--;        }        for(int i=0;i<26;i++)            if(c[i] != 0)                return false;        for(int i=1;i<s1.length();i++)        {            if(isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))                return true;            if(isScramble(s1.substr(0,i), s2.substr(s2.length()-i)) && isScramble(s1.substr(i), s2.substr(0,s2.length()-i)))                return true;               }        return false;    }};

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