小黑怀柔证书下来，腿部酸痛也得到了缓解，跟跑团里的毛毛一起遛龙潭中湖公园，过两天要走闭幕式的leetode之旅的leetcode之旅：18. 四数之和

news2024/12/29 8:55:52

小黑代码(小黑独立做出来)

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        # 数组长度
        n = len(nums)
        if n < 4:
            return []
        # 排序
        nums.sort()
        def three_sum(target_, start, end):
            if end - start < 2:
                return []
            # 结果数组
            res = []
            for i in range(start, end):
                if i != start and nums[i-1] == nums[i]:
                    continue
                # 定义双指针
                head = i + 1
                tail = end
                while head < tail:
                    if nums[head] + nums[tail] < target_-nums[i]:
                        head += 1
                    elif nums[head] + nums[tail] > target_-nums[i]:
                        tail -= 1
                    else:
                        res.append([nums[i], nums[head], nums[tail]])
                        while tail-1 != head and nums[tail] == nums[tail-1]:
                            tail -= 1
                        tail -= 1
            return res
        results = []
        for i in range(n):
            if i != 0 and nums[i-1] == nums[i]:
                continue
            res = three_sum(target-nums[i], i+1, n-1)
            for l in res:
                l.append(nums[i])
                results.append(l)
        return results

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排序 + 双指针

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        # 数组长度
        n = len(nums)
        if n < 4:
            return []
        # 排序数组
        nums.sort()
        # 结果数组
        results = []
        # 开始固定起点，寻找3数之和
        for i in range(n-3):
            # 去除相同元素的情况
            if i != 0 and nums[i-1] == nums[i]:
                continue
            # 剪枝情况
            if nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target:
                break
            if nums[i]+nums[n-3]+nums[n-2]+nums[n-1] < target:
                continue
            # 进行三数之和判断
            for j in range(i+1, n-2):
                # 去除相同元素的情况
                if j != i+1 and nums[j] == nums[j-1]:
                    continue
                # 剪枝情况
                if nums[j]+nums[j+1]+nums[j+2]+nums[i] > target:
                    break
                if nums[j]+nums[n-2]+nums[n-1]+nums[i] < target:
                    continue
                # 定义双指针
                left = j+1
                right = n-1
                while left < right:
                    if nums[left]+nums[right]+nums[i]+nums[j] < target:
                        left += 1
                    elif nums[left]+nums[right]+nums[i]+nums[j] > target:
                        right -= 1
                    else:
                        results.append([nums[left], nums[right], nums[i], nums[j]])
                        while left < right and nums[left] == nums[left+1]:
                            left += 1
                        left += 1
        return results

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数据库练习

620. 有趣的电影

# Write your MySQL query statement below
SELECT
    id, 
    movie,
    description,
    rating
FROM
    cinema
WHERE
    description != 'boring' AND id % 2 = 1
ORDER BY
    rating DESC

Pandas练习

620. 有趣的电影

import pandas as pd

def not_boring_movies(cinema: pd.DataFrame) -> pd.DataFrame:
    # 过滤无聊与id
    data = cinema[(cinema.description!='boring') & (cinema.id%2!=0)]
    data.sort_values('rating', ascending=False, inplace=True)
    return data