Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. （Hard）
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.
分析：
先考虑递归地解法， 对于字符串，枚举其分割位置i - 1，只有有一种分割位置满足
(isScramble(s1.substr(0, i),s2.substr(0, i)) && isScramble(s1.substr(i, s1.size() - i),s2.substr(i, s1.size() - i))
或者
(isScramble(s1.substr(0, i),s2.substr(s1.size() - i, i)) && isScramble(s1.substr(i, s1.size() - i), s2.substr(0,s1.size() - i)))
二者之一成立，则返回true，每个位置枚举后均不成立，则返回false
注：要有一些优化，否则超时过不了，比如s1 == s2 直接返回true， 将s1 s2排个序，之后不相等则直接返回false。
代码：
1 class Solution { 2 public: 3 bool isScramble(string s1, string s2) { 4 if (s1 == s2) { 5 return true; 6 } 7 if (s1.size() != s2.size()) { 8 return false; 9 } 10 if (s1.size() == 1 && s1[0] == s2[0]) { 11 return true; 12 } 13 string cmps1 = s1; 14 string cmps2 = s2; 15 sort(cmps1.begin(), cmps1.end()); 16 sort(cmps2.begin(), cmps2.end()); 17 if (cmps1 != cmps2) { 18 return false; 19 } 20 for (int i = 1; i < s1.size(); ++i) { 21 if ( (isScramble(s1.substr(0, i),s2.substr(0, i)) 22 && isScramble(s1.substr(i, s1.size() - i),s2.substr(i, s1.size() - i))) 23 || (isScramble(s1.substr(0, i),s2.substr(s1.size() - i, i)) 24 && isScramble(s1.substr(i, s1.size() - i), s2.substr(0,s1.size() - i))) ) { 25 return true; 26 } 27 } 28 return false; 29 } 30 };
好像还有三维动态规划的解法，回头再学习一个...