45. Scramble String

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;
    }
};