python 2024-10

语法：while else，continue break 埃氏筛 数的倍数不是素数。

class Solution:
    def countPrimes(self, n: int) -> int:
        res = 0
        for i in range(2, n):
            for j in range(2, i):
                if i % j == 0:
                    break
            else:
                res += 1
        return res

优化

class Solution:
    def countPrimes(self, n: int) -> int:
        num_list = [1] * n
        res = 0
        for i in range(2, n):
            if not num_list[i]: continue
            res += 1
            for j in range(2 * i, n, i):
                num_list[j] = 0

        return res

class Solution:
    def isUgly(self, n: int) -> bool:
        if n <= 0: return False
        factors = [2, 3, 5]
        for fac in factors:
            while n % fac == 0:
                n //= fac
        return n == 1

class Solution:
    def fizzBuzz(self, n: int) -> List[str]:
        res = [''] * 列表包含 n # n 个 空串
        # res = []
        for i in range(1, n + 1):
            if i % 15 == 0: x = "FizzBuzz" # 先处理 15 的倍数
            elif i % 3 == 0: x = "Fizz"
            elif i % 5 == 0: x = "Buzz"
            else: x = str(i) # 转换成字符串
            res[i - 1] = x
            # res.append(x)
        return res

        # return ['Fizz'[i%3*4:]+"Buzz"[i%5*4:] or str(i) for i in range(1, n+1)]

class Solution:
    def findClosestNumber(self, nums: List[int]) -> int:
        ans = inf
        for x in nums:
            if abs(x) == abs(ans): ans = max(x, ans)
            elif abs(x) < abs(ans):  ans = x                
        return ans

标记法

class Solution:
    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
        res = [0, 0]
        n = len(nums)
        i = 0
        for x in nums:
            x %= n
            if nums[x] >= n:
                res[i] = x
                i += 1
                if i == 2: break
            nums[x] += n
        return res

class Solution:
    def hardestWorker(self, n: int, logs: List[List[int]]) -> int:
        ans = pre = most = 0
        for i, t in logs: 
            dif = t - pre
            pre = t
            if most < dif or most == dif and ans > i:
                ans = i
                most = dif

        return ans