中文自我介绍

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英文介绍项目

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Docker用到哪些东西？

请问输出是什么？

python
mylist = [0,1,2,3,4,5,6,7,8]
mylist[-6:-1]

[3, 4, 5, 6, 7]

取name为1111的values的值

name不固定

json
{
    "code": "233",
    "data": [{
            "name": "1111",
            "values": "asdasd"
        }, {
            "name": "2222",
            "values": "dddeee"
        }
    ]
}

jsonpath：$.data[?(@.name == '1111')].values

算法刷题范围和难度选择

根据自己需要选取难易，可按倒序排列从头刷到尾，容易题：261，中等题：566

2023-1-10

454.4Sum II

Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

• 0 <= i, j, k, l < n

• nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1

python
class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        '''
        用字典哈希表，存储前两个数组的两两之和的个数
        再循环后两个数组，如果两两之和的相反数在字典中
        count加相反数在字典中的值，返回count
        '''
        record, count = {}, 0
        for i in nums1:
            for j in nums2:
                if i+j in record:
                    record[i+j] +=1
                else:
                    record[i+j] =1
        for k in nums3:
            for l in nums4:
                if -(k+l) in record:
                    count += record[-(k+l)]
        return count

        
        # record = {}
        # for i in nums1:
        #     for j in nums2:
        #         if i+j in record:
        #             record[i+j] += 1
        #         else:
        #             record[i+j] = 1
        # count = 0
        # for x in nums3:
        #     for y in nums4:
                
        #         if -(x+y) in record:
        #             count += record[-(x+y)]
        # return count

2022-12-01

双指针

python
def maxProfit(prices):
    '''
    121. Best Time to Buy and Sell Stock
    Example 1:
    Input: prices = [7,1,5,3,6,4]
    Output: 5
    Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
	Example 2:
	Input: prices = [7,6,4,3,1]
	Output: 0
	Explanation: In this case, no transactions are done and the max profit = 0.
    先定义最大利润为0，最小价为列表第一位
    循环遍历价格
    定义当前利润等于循环的价格减去最小价
    如果当前利润大于最大利润，替换最大利润为当前利润
    如果当前循环价格小于最小价，替换最小价等于当前循环价格
    循环结束，返回最大利润
    '''
        maxProfit, minPrice= 0, prices[0]
        for price in prices:
            curProfit = price - minPrice
            if curProfit > maxProfit:
                maxProfit = curProfit
            elif price < minPrice:
                minPrice = price
        return maxProfit
    
    '''
    定义最大利润，最小价格，
    循环体内当前最大利润比较，当前最小价格比较
    '''
    minprice, maxprofit = float('inf'), 0
    for i in prices:
        maxprofit = max(maxprofit, i-minprice)
        minprice = min(i, minprice)
    return maxprofit