【leetcode】1417. Reformat The String

题目如下：

Given alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string. 

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Example 4:

Input: s = "covid2019"
Output: "c2o0v1i9d"

Example 5:

Input: s = "ab123"
Output: "1a2b3"

Constraints:

• 1 <= s.length <= 500
• s consists of only lowercase English letters and/or digits.

解题思路：把数字和字母进行分组，如果两者的数量差大于1，说明无法生成符合条件的字符串；如果数字比字母多一个，以数字开头，然后字母轮流交替；如果字母多一个，则以字母开头；如果两者一样，则随便。

代码如下：

class Solution(object):
    def reformat(self, s):
        """
        :type s: str
        :rtype: str
        """
        letters = []
        digits = []
        for i in s:
            if i.isalpha():
                letters.append(i)
            else:
                digits.append(i)
        res = ''
        if abs(len(letters) - len(digits)) >= 2:
            return ''
        elif len(letters) - len(digits) == 1:
            while len(letters) > 0 or len(digits) > 0:
                res += letters.pop(0)
                if len(digits) > 0:
                    res += digits.pop(0)
        elif len(digits) - len(letters) == 1:
            while len(digits) > 0 or len(letters) > 0:
                res += digits.pop(0)
                if len(letters) > 0:
                    res += letters.pop(0)
        else:
            while len(digits) > 0 or len(letters) > 0:
                res += digits.pop(0)
                res += letters.pop(0)
        return res