小黑受到了未来的焦虑，周四继续参加团跑活动仰山跑，跑奥森的坡，越跑越上瘾更加热爱生活的leetcode之旅:LCR 008. 长度最小的子数组

小黑代码1

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        # 数组长度
        n = len(nums)
        # 双指针
        head = 0
        tail = 0
        # 中间变量
        sum_ = 0
        # 结果变量
        res = n+1
        # 开始双指针迭代
        while tail < n:
            sum_ += nums[tail]
            tail += 1
            while sum_ >= target:
                if tail - head < res:
                    res = tail - head
                sum_ -= nums[head]
                head += 1
        return res if res != n+1 else 0

小黑代码2

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        # 数组长度
        n = len(nums)
        # 构造累加数组
        arr = [0]
        for i in range(n):
            arr.append(arr[i]+nums[i])
        if arr[-1] < target:
            return 0
        print(arr)
        # 结果变量
        res = n+1
        # 对每个元素进行二分查找
        for i in range(1, n+1):
            if arr[i] < target:
                continue 
            t = self.bin_search(arr, 0, i-1, arr[i]-target)
            if arr[t] > arr[i]-target:
                t -= 1
            if i-t < res:
                print(t, i)
                res = i-t
        return res
    def bin_search(self, arr, start, end, target):
        while start <= end:
            mid = (start + end) // 2
            if arr[mid] >= target:
                end = mid - 1
            else:
                start = mid + 1
        return start

二分法

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        # 数组长度
        n = len(nums)
        # 构造累加数组
        arr = [0]
        for i in range(n):
            arr.append(arr[i]+nums[i])
        if arr[-1] < target:
            return 0
        # 结果变量
        res = n+1
        # 对每个元素进行二分查找
        for i in range(n):
            end = self.bin_search(arr, i+1, n, arr[i]+target)
            if end <= n:
                res = min(res, end-i)
        return res
    def bin_search(self, arr, start, end, target):
        while start <= end:
            mid = (start + end) // 2
            if arr[mid] >= target:
                end = mid - 1
            else:
                start = mid + 1
        return start

数据库练习

570. 至少有5名直接下属的经理（小黑独立做出）

by 小黑

# Write your MySQL query statement below
SELECT 
    name
FROM
    Employee
WHERE
    id  IN
        (SELECT
            managerId
        FROM
            Employee
        WHERE
            managerId IS NOT NULL
        GROUP BY
            managerId
        HAVING
            COUNT(name) >= 5)

# Write your MySQL query statement below
SELECT
    e1.name
FROM
    Employee e1 INNER JOIN
        (SELECT
            managerId
        FROM
            Employee
        WHERE
            managerId IS NOT NULL
        GROUP BY
            managerId
        HAVING 
            COUNT(*) >= 5 
        ) AS e2 ON e1.id = e2.managerId

pandas练习

import pandas as pd

def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
    # 分组，求和
    manage_data = employee.groupby('managerId').size().reset_index(name='count')
    # 筛选符合条件的数据
    manage_id = list(manage_data[manage_data['count']>=5]['managerId'])
    # 通过manageId找到对应姓名

    bool_ = employee['id'].apply(lambda x:x in manage_id)
    return pd.DataFrame(data={'name':employee['name'][bool_].tolist()}, columns=['name'])

小黑生活

跑量少的可怜，工作以后要调整自己鸭

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