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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.
这道题，题目都没读懂是什么意思~~，为自己的理解能力捉急~~，参考了http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/和http://blog.theliuy.com/scramble-string/这两篇博客，看了代码才懂题的意思。用的就是动态规划。
c++代码：
#include <string>
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int char_size;
bool isScramble(string s1, string s2) {
char_size = 26;
return isScrambleHelper(s1, s2);
}
bool isScrambleHelper(string &s1, string &s2) {
if (s1.size() != s2.size())
return false;
if (s1 == s2)
return true;
int size = s1.size();
vector<int> bucket(char_size, 0);
string s11, s12, s21, s22;
// Check wheter they have the same chars
for (int i = 0; i < s1.size(); ++i) {
bucket[s1[i] - ‘a‘] += 1;
bucket[s2[i] - ‘a‘] -= 1;
}
for (int i = 0; i < char_size; ++i) {
if (bucket[i] != 0)
return false;
}
for (int i = 1; i < size; ++i) {
s11 = s1.substr(0, i);
s12 = s1.substr(i);
s21 = s2.substr(0, i);
s22 = s2.substr(i);
if (isScrambleHelper(s11, s21) && isScrambleHelper(s12, s22))
return true;
s21 = s2.substr(size - i);
s22 = s2.substr(0, size - i);
if (isScrambleHelper(s11, s21) && isScrambleHelper(s12, s22))
return true;
}
return false;
}
};
int main()
{
Solution s = Solution();
bool r = s.isScramble("great", "rgeat");
cout << r << endl;
}
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原文地址：http://www.cnblogs.com/zhutianpeng/p/4251539.html
