leetcode937. Reorder Log Files

Posted 2021-01-19 seyjs

题目如下：

You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier.  Then, either:

• Each word after the identifier will consist only of lowercase letters, or;
• Each word after the identifier will consist only of digits.

We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

Return the final order of the logs.

 

Example 1:

Input: ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]

Note:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

解题思路：题目实在太简单了，我的方法是创建两个数组letter和digit，接下来遍历logs，如果logs[i]的最后一个字符是数字，存入digit；否则，存入letter。遍历完成后，对letter进行排序，最后返回letter + digit。

随便说说：最近真的是太忙了，基本没有时间做题。

代码如下：

class Solution(object):
    def reorderLogFiles(self, logs):
        """
        :type logs: List[str]
        :rtype: List[str]
        """
        letter = []
        digit = []
        for i in logs:
            if i[-1].isdigit():
                digit.append(i)
            else:
                letter.append(i)
        def cmpf(v1,v2):
            lv1 = v1.split(‘ ‘)
            lv2 = v2.split(‘ ‘)
            for i in range(1,min(len(lv1),len(lv2))):
                if lv1[i] == lv2[i]:
                    continue
                return cmp(lv1[i],lv2[i])
            return len(lv1) - len(lv2)

        letter.sort(cmp = cmpf)
        return letter + digit