LeetCode 笔记系列 19 Scramble String [合理使用递归]

题目：

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.

完全没有思路。

卡了很久，最后参考这里的解释。写了一个递归的版本解决。

代码：

public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
       if(!isContainSameChars(s1, s2))return false;
       if(s1.equals(s2)) return true;
       for(int split = 1; split < s1.length(); split++){
           String s11 = s1.substring(0, split);
           String s12 = s1.substring(split);
           
           String s21 = s2.substring(0, split);
           String s22 = s2.substring(split);
           if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
           
           s21 = s2.substring(0, s2.length() - split);
           s22 = s2.substring(s2.length() - split);
           if(isScramble(s11, s22) && isScramble(s12, s21)) return true;
       }
       return false;
    }

View Code

其实不算难。除非你想不到递归。