[LeetCode] 87. Scramble String

不会翻译这个标题。。但是题意是给一个string s1，这个string被表示成一个二叉树。在树的叶节点，每个字母会被分到二叉树的每一个最小的分支上自成一支。

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".

如上例子，great和rgeat是互为scramble string的。这题给的tag是DP但是DP的实现是三维的，几乎不可能在面试过程中完美实现。这里我给出的是递归解法，比较直观。

时间O(n!) - 我其实不是很确定

空间O(n) - 因为递归用到栈空间

 1 /**
 2  * @param {string} s1
 3  * @param {string} s2
 4  * @return {boolean}
 5  */
 6 var isScramble = function(s1, s2) {
 7     if (s1 === s2) return true;
 8     const letters = new Array(128).fill(0);
 9     const a = 'a'.charCodeAt(0);
10     for (let i = 0; i < s1.length; i++) {
11         letters[s1.charCodeAt(i) - a]++;
12         letters[s2.charCodeAt(i) - a]--;
13     }
14     for (let i = 0; i < 128; i++) {
15         if (letters[i] !== 0) return false;
16     }
17     for (let i = 1; i < s1.length; i++) {
18         if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i)))
19             return true;
20         if (
21             isScramble(s1.substring(0, i), s2.substring(s2.length - i)) &&
22             isScramble(s1.substring(i), s2.substring(0, s2.length - i))
23         )
24             return true;
25     }
26     return false;
27 };