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.
给定一个字符串s1, 我们可以将其分裂表示成一颗二叉树，通过调换二叉树中非空节点的左右孩子，我们可以组合得到新的字符串s2, 我们称s2是s1的scrambled string。
class Solution {
public:
bool isValid(string s1, string s2){
//s1和s2应该有相同种的字符，且各字符出现的次数相同
sort(s1.begin(), s1.end());
sort(s2.begin(), s2.end());
if(s1.compare(s2)==0)return true;
return false;
}
bool isScramble(string s1, string s2) {
if(s1.length()!=s2.length())return false;
if(s1.length()==0 || s2.length()==0)return false;
if(s1.compare(s2)==0)return true; //如果s1和s2相同，则返回true;
if(s1.length()==1)return false; //如果s1和s2长度为1， 且不相等，则返回false
//考虑所有的分裂情况，然后递归判断
for(int len=1; len<=s1.length()-1; len++){
string s1left = s1.substr(0, len);
string s1right = s1.substr(len, s1.length()-len);
string s2left = s2.substr(0, len);
string s2right = s2.substr(len, s1.length()-len);
//在进行下一轮递归是需要先判断输入元素是否合法，如果不合法就没必要进行递归，过多的递归会使时间成本陡增
if(isValid(s1left, s2left) && isScramble(s1left, s2left) && isScramble(s1right, s2right))return true;
s2left = s2.substr(0, s1.length()-len);
s2right = s2.substr(s1.length()-len, len);
if(isValid(s1left, s2right) && isScramble(s1left, s2right) && isScramble(s1right, s2left))return true;
}
return false;
}
};LeetCode: Scramble String [87],布布扣,bubuko.com
LeetCode: Scramble String [87]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/27795553
