[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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

扰乱字符串。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。

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

JavaScript实现

/**
 * @param {string} s1
 * @param {string} s2
 * @return {boolean}
 */
var isScramble = function(s1, s2) {
    if (s1 === s2) return true;
    const letters = new Array(128).fill(0);
    const a = 'a'.charCodeAt(0);
    for (let i = 0; i < s1.length; i++) {
        letters[s1.charCodeAt(i) - a]++;
        letters[s2.charCodeAt(i) - a]--;
    }
    for (let i = 0; i < 128; i++) {
        if (letters[i] !== 0) return false;
    }
    for (let i = 1; i < s1.length; i++) {
        if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i)))
            return true;
        if (
            isScramble(s1.substring(0, i), s2.substring(s2.length - i)) &&
            isScramble(s1.substring(i), s2.substring(0, s2.length - i))
        )
            return true;
    }
    return false;
};

Java实现 - 超时

class Solution {
    public boolean isScramble(String s1, String s2) {
        // corner case
        if (s1 == null || s2 == null) {
            return false;
        }
        if (s1.equals(s2)) {
            return true;
        }
        // normal case
        int[] letters = new int[26];
        int len = s1.length();
        for (int i = 0; i < len; i++) {
            letters[s1.charAt(i) - 'a']++;
            letters[s2.charAt(i) - 'a']--;
        }
        for (int i = 0; i < 26; i++) {
            if (letters[i] != 0) {
                return false;
            }
        }
        for (int i = 1; i < len; i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) {
                return true;
            }
            if (isScramble(s1.substring(0, i), s2.substring(len - i)) && isScramble(s1.substring(i), s2.substring(0, len - i))) {
                return true;
            }
        }
        return false;
    }
}

Java另一种带memorization的实现

class Solution {
    HashMap<String, Boolean> map = new HashMap<>();

    public boolean isScramble(String s1, String s2) {
        StringBuilder sb = new StringBuilder();
        sb.append(s1);
        sb.append(s2);
        String key = sb.toString();

        if (map.containsKey(key)) {
            return map.get(key);
        }

        if (s1.equals(s2)) {
            map.put(key, true);
            return true;
        }

        int[] letters = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            letters[s1.charAt(i) - 'a']++;
            letters[s2.charAt(i) - 'a']--;
        }
        for (int i = 0; i < 26; i++) {
            if (letters[i] != 0) {
                map.put(key, false);
                return false;
            }
        }

        for (int i = 1; i < s1.length(); i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) {
                map.put(key, true);
                return true;
            }

            if (isScramble(s1.substring(0, i), s2.substring(s1.length() - i))
                    && isScramble(s1.substring(i), s2.substring(0, s1.length() - i))) {
                map.put(key, true);
                return true;
            }
        }

        map.put(key, false);
        return false;
    }
}

LeetCode 题目总结