136. Single Number

Given an 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?

 

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dic = {}
        for num in nums:
            dic[num] = dic.get(num, 0)+1
        for key, val in dic.items():
            if val == 1:
                return key 

dic.get(num,0) 对键num来说，如果它出现在字典里，返回它对应的值；如果没有，返回default定义的值

dic.items()返回键值对

 

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        for num in nums:
            res ^= num
        return res 

^=按位异或

 

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return 2*sum(set(nums))-sum(nums)

set()变成集合，集合是无序不重复集

 

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return reduce(lambda x, y: x ^ y, nums)

python中的reduce内建函数是一个二元操作函数，他用来将一个数据集合（链表，元组等）中的所有数据进行下列操作：用传给reduce中的函数 func()（必须是一个二元操作函数）先对集合中的第1，2个数据进行操作，得到的结果再与第三个数据用func()函数运算，最后得到一个结果。

 

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return reduce(operator.xor, nums)