Re: [閒聊] 每日leetcode

[sustainer123 (caster )]

※ 引述《DJYOMIYAHINA (通通打死)》之銘言：
: 今天本肥肥也是個稱職的ez守門員
: 準備出門當帕魯
: def minOperations(self, logs: List[str]) -> int:
:     base = 0
:     for log in logs:
:         if log == "../":
:             base = max(base-1, 0)
:         elif log != "./":
:             base += 1
:     return base


https://leetcode.com/problems/crawler-log-folder

1598. Crawler Log Folder

有一log紀錄使用者更改資料夾的操作

操作有三:

1 ../:回到父資料夾 假如已在主資料夾 則不變

2 ./:不變

3 x/:前往子資料夾x


請回傳完成log的資料夾操作後 回到主資料夾的最小操作



Example 1:



Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use this change folder operation "../" 2 times and go back to
the main folder.
Example 2:



Input: logs = ["d1/","d2/","./","d3/","../","d31/"]
Output: 3
Example 3:

Input: logs = ["d1/","../","../","../"]
Output: 0


Constraints:

1 <= logs.length <= 103
2 <= logs[i].length <= 10
logs[i] contains lowercase English letters, digits, '.', and '/'.
logs[i] follows the format described in the statement.
Folder names consist of lowercase English letters and digits.


思路:
照題目需求模擬


Python Code:
class Solution:
    def minOperations(self, logs: List[str]) -> int:
        result = 0

        for log in logs:
            if log == "../":
                if result > 0:
                    result -= 1
                else:
                    continue
            elif log == "./":
                continue
            else:
                result += 1

        return result


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