Scramble String
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.
注：Google 面试题。
分析： 从上往下考虑，比然存在一个位置，使得两个字符串分成相同的两个子串，他们位置前后相同或者前后相反。如此递归即可。能递归也就能动态规划记住子问题的解。但是本题没有使用动态规划也很快AC.
bool hasSameAlpha(string &s1, string &s2) {
int v = 0;
for(int i = 0; i < s1.size(); ++i) v ^= (s1[i] ^ s2[i]);
return v == 0;
}
class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1.size() != s2.size()) return false;
if(s1 == s2) return true;
if(!hasSameAlpha(s1, s2)) return false;
int n = s1.size();
for(int i = 1; i < n; ++i) {
string s11 = s1.substr(0, i);
string s12 = s1.substr(i);
string s21 = s2.substr(0, n-i);
string s22 = s2.substr(n-i);
if((isScramble(s11, s2.substr(0, i)) && isScramble(s12, s2.substr(i))) || (isScramble(s11, s22) && isScramble(s12, s21)))
return true;
}
return false;
}
};