本文共 810 字，大约阅读时间需要 2 分钟。

给定一个非空的整数数组,除了某个元素只出现一次以外其余每个元素均出现了两次.找出那个只出现了一次的元素

说明 你的算法应该具有线性时间复杂度,你可以不用额外的空间来实现么

示例 1

输入 [2, 2, 1,]

输出 1

示例2

输入 [4, 1, 2, 1, 2]

输出 4

首先给你数组

r_number = [1, 2, 3, 2, 3, 6, 1, ]

耗时比较长的思路,双层循环,将第一个值取出来与列表剩下的值相对比较如果有相等的就跳出循环然后利用elese弹出该下标的值

r_number = [1, 2, 3, 2, 3, 6, 1,]c = []for i in range(len(r_number)):	for k in range(len(r_number)):		if r_number[i] == r_number[k]:			breakelse:	print(r_number[i])

第三个就是利用集合…

r_number = [1, 2, 3, 2, 3, 6, 1,]c = 2 * sum(set(r_number)) - sum(r_number)

第四个就是按照切片…这个不是我想出来的

r_number = [1, 2, 3, 2, 3, 6, 1,]for i in range(len(r_number)):	if r_number[i] not in r_number[0:i] and r_number[i] not in	r_number[i+1 : ]:		print(r_number[i])

利用HASH进行判断, L神设计

nums = [1,2,3,4,5,6,7,8,9,1,2,3,5,6,7,8,9]dic = {}for i in nums:	dic[i] = dic.get(i,0) + 1for i in dic:	if dic[i] == 1: