Re: [閒聊] 每日leetcode

[JerryChungYC (JerryChung)]

https://leetcode.com/problems/find-missing-observations

2028. Find Missing Observations

※ 引述《sustainer123 (caster )》之銘言：
: ※ 引述《DJYOMIYAHINA (通通打死)》之銘言：
: : 有一大坨比我快十倍的不知道怎麼用的==
: : 但我看答案都差不多ㄚ
: : 還是有改測資
: : def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
: :     m = len(rolls)
: :     summation = mean*(n+m)
: :     sum_n = summation - sum(rolls)
: :     if sum_n<n or sum_n>6*n:
: :         return []
: :     res = sum_n-(sum_n//n)*n
: :     ans = [(sum_n//n)+1] * res + [(sum_n//n)] * (n-res)
: :     return ans
: 思路:
: 先求(n+m)總和跟m總和的差值
: 然後 求n的平均 餘數後面慢慢加上去
: 另外 裡面的元素限定1-6 所以就加個if判斷
: Python Code:
: class Solution:
:     def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
:         diff = mean * (len(rolls) + n) - sum(rolls)
:         diff_avg = diff // n
:         space = (6 - diff_avg) * n
:         remainder = diff % n
:         result = []
:         if diff < n or diff > 6 * n or remainder > space:
:             return result
:         for _ in range(n):
:             if remainder > 6 - diff_avg:
:                 remainder -= 6 - diff_avg
:                 result.append(6)
:             else:
:                 result.append(diff_avg + remainder)
:                 remainder = 0
:         return result
: 寫得狗幹醜 我就這樣了

思路：差不多
先算出差多少數
如果在 n 跟 6n 之外就回傳 []
再算平均與餘數 看幾個餘數就加幾個

Python Code:

class Solution:
    def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
        lose = mean * (len(rolls) + n) - sum(rolls)
        if 1 * n > lose or lose > 6 * n: return []
        avg, remainder = divmod(lose, n)
        return [avg] * (n - remainder) + [avg+1] * remainder

你們寫的都好厲害 :(

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推 dont: leetcode應該是只算aux space