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
思路：
- 先排序
- 然后蛮力法
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums.sort()
print(nums)
if len(nums) == 1:
return nums[0]
if nums[0] != nums[1]:
return nums[0]
if nums[len(nums)-1] != nums[len(nums)-2]:
return nums[len(nums)-1]
for i in range(1, len(nums)-2):
if nums[i] != nums[i+1] and nums[i] != nums[i-1]:
return nums[i]
break
if __name__ == "__main__":
s = [1]
print(Solution().singleNumber(s))
