LeetCode-87 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.

 

思路：

1. 如果只有一个s1.length == s2.length == 1;则只需判断s1 == s2;

2. 如果s1.length == s2.length == 2；则需要判断(s1[0] == s2[0] && s1[1]==s2[1]) || (s1[0] == s2[1] && s1[1] == s2[0])

3. 如果s1.length == s2.length == 3;则需要判断

(s1.substring(0, i) == s2.substring(0, i) && s1.substring(i) == s2.substring(i)) || 

(s1.substring(0, i) == s2.substring(s2.length()-i) && s1.substring(i) == s2.substring(0, s2.length()-i))

如果仅依靠以上的判断，需要花费较多的时间，因此需要添加其他判断条件来减少递归次数。

对于这道题，必要的条件是s1与s2的长度相同，且s1与s2的字符集相同。

 

代码如下:

public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length()) {
            return false;
        }
        if(s1.length() == 1) {
            return s1.equals(s2);
        }
        int charset[] = new int[26];
        for(int i=0; i<s1.length(); i++) {
            charset[s1.charAt(i) - 'a']++;
            charset[s2.charAt(i) - 'a']--;
        }
        for(int i=0; i<26; i++) {
            if(charset[i] != 0)
                return false;
        }
        
        for(int i=1; i<s1.length(); i++) {
            boolean result = ( isScramble(s1.substring(0, i), s2.substring(0, i)) && 
                                isScramble(s1.substring(i), s2.substring(i)) ) ||
                             (isScramble(s1.substring(0, i), s2.substring(s2.length()-i)) && 
                                isScramble(s1.substring(i), s2.substring(0, s2.length()-i)));
                    
            if(result) {
                return true;
            }
        }
        return false;
    }

 

 

对于本题，还可以通过动态规划的解法来提高效率。之后会更新。