題目如下：

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:

0 <= logs.length <= 100

3 <= logs[i].length <= 100

logs[i]?is guaranteed to have an identifier, and a word after the identifier.

解題思路：題目實在太簡單了，我的方法是創建兩個數組letter和digit，接下來遍歷logs，如果logs[i]的最后一個字符是數字，存入digit；否則，存入letter。遍歷完成后，對letter進行排序，最后返回letter + digit。

隨便說說：最近真的是太忙了，基本沒有時間做題。

代碼如下：

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]:continuereturn cmp(lv1[i],lv2[i])return len(lv1) - len(lv2)letter.sort(cmp = cmpf)return letter + digit

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轉載于:https://www.cnblogs.com/seyjs/p/9958017.html

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