第 273 场周赛

文章目录

      • 题目1
        • 代码
        • 运行结果
      • 题目2
        • 代码
        • 运行结果
      • 题目3
        • 代码
        • 运行结果
      • 总结

题目1

'''
Description: 5963. 反转两次的数字
Autor: 365JHWZGo
Date: 2021-12-26 10:33:25
LastEditors: 365JHWZGo
LastEditTime: 2021-12-26 10:36:27
Author: localhost
'''

代码

class Solution(object):
    def isSameAfterReversals(self, num):
        """
        :type num: int
        :rtype: bool
        """
        if num == 0:
            return True
        l = list(str(num))
        if int(l[-1]) == 0:
            return False
        else:
            return True

运行结果

在这里插入图片描述

题目2

'''
Description: 5964. 执行所有后缀指令
Autor: 365JHWZGo
Date: 2021-12-26 10:42:03
LastEditors: 365JHWZGo
LastEditTime: 2021-12-26 14:05:10
Author: localhost
'''

代码

class Solution(object):
    def executeInstructions(self, n, startPos, s):
        """
        :type n: int
        :type startPos: List[int]
        :type s: str
        :rtype: List[int]
        """
        x,y = startPos
        res = []
        
        for i in range(len(s)):
            step = 0
            for j in range(i,len(s)):
                if s[j] == 'R':
                    y = y+1
                elif s[j] == 'L':
                    y = y-1
                elif s[j] == 'U':
                    x = x-1
                else:
                    x = x+1
                if x>=n or y>=n or x<0 or y <0:
                    break
                else:
                    step +=1
            res.append(step)
            x,y = startPos
        print(res)
        return res

运行结果

第 273 场周赛_第1张图片

题目3

'''
Description: 5965. 相同元素的间隔之和
Autor: 365JHWZGo
Date: 2021-12-26 10:42:03
LastEditors: 365JHWZGo
LastEditTime: 2021-12-26 14:02:10
Author: localhost
'''

代码


class Solution(object):
    def getDistances(self, arr):
        """
        :type arr: List[int]
        :rtype: List[int]
        """
        res = [0 for _ in range(len(arr))]
        dp = [[] for _ in range(100000)]
        # print(dp)
        #将所有相同的元素的下标放在一起
        for i in range(len(arr)):
            dp[arr[i]-1].append(i)

        
        l = list(set(arr))
        for i in l:
        	#resf,前缀和,resb,后缀和
            resf,resb = 0,0
            #计算前缀和
            for j in range(1, len(dp[i-1])):
                resf = resf+(dp[i-1][j]-dp[i-1][j-1])*j
                res[dp[i-1][j]] += resf
			#计算后缀和
            for j in range(len(dp[i-1])-2,-1,-1):
                resb = resb+(dp[i-1][j+1]-dp[i-1][j])*(len(dp[i-1])-1-j)
                res[dp[i-1][j]] += resb
        return res

运行结果

第 273 场周赛_第2张图片

总结

前两道题不难,第三道题也不难,但是有时间限制,你必须要降低复杂度才可以
一开始我写的是

class Solution(object):
    def getDistances(self, arr):
        """
        :type arr: List[int]
        :rtype: List[int]
        """
        dp = [[0 for _ in range(len(arr))]for _ in range(len(arr))]
        for i in range(len(arr)):
            for j in range(i+1,len(arr)):
                if arr[i] == arr[j]:
                    dp[i][i]+=j-i
                    dp[j][j]+=j-i
        res = [dp[i][i] for i in range(len(arr))]
        return res

其实就是依次遍历相加距离,奈何超时间了
第 273 场周赛_第3张图片
后来又进行了改进相当于我只遍历当前为arr[i]的那些元素所在下标,奈何题目就偏要让我再降复杂度~

class Solution(object):
    def getDistances(self, arr):
        """
        :type arr: List[int]
        :rtype: List[int]
        """
        res = [0 for _ in  range(len(arr))]
        dp = [[] for _ in range(100000)]
        # print(dp)
        for i in range(len(arr)):
            dp[arr[i]-1].append(i)
        
        l = list(set(arr))
        for i in l:
            for j in range(len(dp[i-1])-1):
                for k in range(j+1,len(dp[i-1])):
                    res[dp[i-1][j]] += dp[i-1][k]-dp[i-1][j]
                    res[dp[i-1][k]] += dp[i-1][k]-dp[i-1][j]
        return res

在这里插入图片描述

再接再厉!下周争取多做出一道题。

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