Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4
class Solution:
singleNumber(self, nums):
for i in nums:
a = nums.count(i) # list내의 숫자의 개수
if a == 1: # 숫자의 개수가 1이면 결과값 반환
return i
class Solution:
singleNumber(self, nums):
a = 0
for i in nums:
a ^= i
return a