[LeetCode] Scramble String 解题报告

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.

» Solve this problem

[解题思路]
首先想到的是递归，简单明了，对两个string进行partition，然后比较四个字符串段。但是递归的话，这个时间复杂度比较高。然后想到能否DP，但是即使用DP的话，也要O(n^3)。想想算了，还是在递归里做些剪枝，这样就可以避免冗余计算：

• 对于每两个要比较的partition，统计他们字符出现次数，如果不相等返回。

 上网搜了一下，有人说可以O(n)做出来。http://www.mitbbs.com/article_t/JobHunting/32114513.html
但是我没办法证明他这个算法的正确性。感觉这道题应该有O(n)的解法，一时想不出来。

1:       bool isScramble(string s1, string s2) {   
2:            // Start typing your C/C++ solution below   
3:            // DO NOT write int main() function   
4:            if(s1.size() != s2.size()) return false;   
5:            int A[26];   
6:            memset(A,0,26*sizeof(A[0]));   
7:            for(int i =0;i<s1.size(); i++)   
8:            {   
9:                 A[s1[i]-'a']++;   
10:            }   
11:            for(int i =0;i<s2.size(); i++)   
12:            {   
13:                 A[s2[i]-'a']--;   
14:            }   
15:            for(int i =0;i<26; i++)   
16:            {   
17:                 if(A[i] !=0)   
18:                 return false;   
19:            }   
20:            if(s1.size() ==1 && s2.size() ==1) return true;   
21:            for(int i =1; i< s1.size(); i++)   
22:            {   
23:                 bool result= isScramble(s1.substr(0, i), s2.substr(0, i))   
24:                      && isScramble(s1.substr(i, s1.size()-i), s2.substr(i, s1.size()-i));   
25:                 result = result || (isScramble(s1.substr(0, i), s2.substr(s2.size() - i, i))   
26:                      && isScramble(s1.substr(i, s1.size()-i), s2.substr(0, s1.size()-i)));   
27:                 if(result) return true;   
28:            }   
29:            return false;   
30:       }