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。