代码(Python3) class Solution: def findDuplicate(self, nums: List[int]) -> int: # 二分区间左边界,初始化为 1 l: int = 1 # 二分区间右边界,初始化为 n r: int = len(nums) - 1 # 当前区间不为空时,继续二分 while l <= r: # 计算区间中点 mid mid: int = (l + r) >> 1 #...
14. We were discussing how to duplicate a random number generation process in Python. It was like trying to recreate a magic trick. We had to understand the underlying algorithm of the random module. 我们正在讨论如何在Python中复制一个随机数生成过程。这就像试图重新创造一个魔术。我们必须理解随机...
假设存在不包含key的环,起点0在不包含key的环中绕圈。 代码如下 classSolution:deffindDuplicate(self, nums:List[int]) ->int:# 如果只有两个元素,第一个元素一定是重复元素iflen(nums) ==2:returnnums[0]# fast每次走两步,slow每次走一步,起始点可以为任意位置fast =0slow =0# python没有do while,所以...
Given an array of integersnumscontainingn + 1integers where each integer is in the range[1, n]inclusive. There is only one duplicate number innums, returnthis duplicate number. Follow-ups: How can we prove that at least one duplicate number must exist innums? Can you solve the problem wi...
place_of_first_number = a.index(value) a[place_of_first_number] = 'string' place_of_second_number = a.index(value) if place_of_second_number < starting_distance: starting_distance = place_of_second_number answer = value a=b n+=1 if n == len(a)-1: return answer return answer...
Your runtime complexity should be less thanO(n2). There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一...
How to remove duplicate number python3 21st Oct 2018, 9:04 AM Goodboy08 + 3 If you mean from a list, just transform it into a set: k=[1, 1,1,1,15] s=set(k) 21st Oct 2018, 9:07 AM fra + 2 If the duplicates are in a list, and you want to remove duplicates while ...
initially we fill the tree set (red black tree) with first k numbers// after k characters we need to delete the first number, because we move the window// according the requirements |i-j| <= kif (i > k) {set.remove(nums[i - k - 1]);}// if tree has already such number, ...
Checking whether a number has appeared can be done not only with a BST but also with a hash table. This would improve the time complexity, but the constant factors can be very large. In this case, the complexity of the second method would become O(n⋅kk−1)O(n⋅kk−1). ...
Python Code: # Define a function 'unique_list' that removes duplicates from a listdefunique_list(l):# Create an empty list 'temp' to store unique elementstemp=[]# Iterate through the elements of the input list 'l'forxinl:# Check if the element 'x' is not in the 'temp' list (i....