Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

    def singleNumber(self, A):
l = len(A)
if l < 2:
return A[0]
A.sort()
for i in range(0,l-1,2):
if A[i] != A[i+1]:
return A[i]
return A[l-1]

思路:先排序,再找结果。

注意:当A的长度为1时,以及结果为最后一个时(当时没处理最后一个数字为结果导致出现错误,for循环中将最后一个排除了,所以后面要做处理)。

Single Number II

 

Given an array of integers, every element appears three times except for one. Find that single one.

只要将range里面的步长改为3就行了。

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