leetcode -- 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.

 

[解题思路]

DFS+Pruning

DFS的终止条件，以及使用pruning来减少不必要的比较

public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len1 = s1.length(), len2 = s2.length();
        if(len1 != len2){
            return false;
        }
        
        int[] count = new int[26];
        for(int i = 0; i < len1; i++){
            count[s1.charAt(i) - 'a'] ++;
        }
        
        for(int i = 0; i < len2; i++){
            count[s2.charAt(i) - 'a'] --;
        }
        
        for(int i = 0; i < 26; i++){
            if(count[i] != 0){
                return false;
            }
        }
        
        if(len1 == 1 && len2 == 1){
            return true;
        }
        
        for(int i = 1; i < len1; i++){
            boolean result = isScramble(s1.substring(0,i), s2.substring(0, i)) &&
                isScramble(s1.substring(i), s2.substring(i));
            
            result = result || (isScramble(s1.substring(0, i), s2.substring(len2 - i)) && 
                    isScramble(s1.substring(i), s2.substring(0, len2 - i)));
            
            if(result){
                return result;
            }
        }
        return false;
    }

2.DP

 

http://blog.sina.com.cn/s/blog_b9285de20101gy6n.html

http://fisherlei.blogspot.com/2013/01/leetcode-scramble-string.html

http://tech-wonderland.net/blog/leetcode-scramble-string.html