题目连接：Leetcode 087 Scramble String
解题思路：DFS，每次枚举分割点，判断左右子串是否是Scramble String。不加剪枝的话是会超时，在每次递归判断前，判断当前两个字符串字符个数是否一致，不一致的话直接返回false。
class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
int n = s1.size();
int s1Map[128] = {0}, s2Map[128] = {0};
for (int i = 0; i < n; i++) {
s1Map[s1[i]]++;
s2Map[s2[i]]++;
}
for (int i = 0; i < 128; i++) if(s1Map[i] != s2Map[i])
return false;
for (int i = 1; i < n; 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(n-i))
&& isScramble(s1.substr(i), s2.substr(0, n-i)) )
return true;
}
return false;
}
};