2164. 对奇偶下标分别排序

拆分+重排，一行搞定

class Solution:
    def sortEvenOdd(self, nums: List[int]) -> List[int]:
        return [*chain(*zip(sorted(nums[::2]),sorted(nums[1::2],reverse=True)),([max(nums[::2])] if len(nums)&1 else []))]

2165. 重排数字的最小值

分正负，利用字符串重排

class Solution:
    def smallestNumber(self, num: int) -> int:
        seq,n=sorted([x for x in str(num) if x not in ('0','-')]),str(num).count('0')
        return int(seq[0]+'0'*n+''.join(seq[1:])) if num>0 else -int(''.join(seq[::-1])+'0'*n)

2166. 设计位集

要求：设计O(1)的flip()（因为flip最多可能调用 $10^5$ 次）
提示：至多调用 toString 方法 5 次（说明toString()可以为O(n)）
结论：设置标记位flag，只在toString()中翻转（而在其他函数中只修改标记位）

class Bitset:

    def __init__(self, size: int):
        self.num=[0]*size
        self.cnt=0
        self.size=size
        self.flag=False


    def fix(self, idx: int) -> None:
        if self.num[idx] ^ self.flag == 0:
            self.cnt+=1
            self.num[idx] = 1-self.num[idx]


    def unfix(self, idx: int) -> None:
        if self.num[idx] ^ self.flag == 1:
            self.cnt-=1
            self.num[idx] = 1- self.num[idx]


    def flip(self) -> None:
        self.flag = not self.flag
        self.cnt = self.size - self.cnt



    def all(self) -> bool:
        return self.cnt==self.size


    def one(self) -> bool:
        return self.cnt>0


    def count(self) -> int:
        return self.cnt


    def toString(self) -> str:
        return ''.join(map(str,[i^self.flag for i in self.num]))



# Your Bitset object will be instantiated and called as such:
# obj = Bitset(size)
# obj.fix(idx)
# obj.unfix(idx)
# obj.flip()
# param_4 = obj.all()
# param_5 = obj.one()
# param_6 = obj.count()
# param_7 = obj.toString()

2167. 移除所有载有违禁货物车厢所需的最少时间

从排行榜看到的大神一行解法👇

class Solution:
    def minimumTime(self, s):
        return min(reduce(lambda r, t: (min(r[0], r[1] + n - t[0]), min(r[1] + [0, 2][t[1] == '1'], t[0] + 1)), enumerate(s), (n := len(s), 0)))

很巧妙的利用了reduce，其实等同于下面代码：

class Solution:
    def minimumTime(self, s):
        n=len(s)
        res=(n,0)
        for i,j in enumerate(s):
            res=(min(res[0], res[1] + n - i), min(res[1] + [0, 2][j == '1'], i + 1))
        return min(res)