Single Number

标准异或的定义，python利用reduce函数一行就可以搞定了

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

Single Number II

这题有两种做法，都不是很记得解题思路了，先把代码写下来，下面这种是数三出现的次数，比较好理解

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res = 0
        for i in xrange(32):
            base = 1 << i
            count = 0
            for num in nums:
                if num & base:
                    count += 1

            if count % 3:
                res |= base
        return res if abs(res) < abs(2 ** 32 - res) else res - 2 ** 32

另外还有bit位操作的方法，完全不知道是什么原理了

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a0, a1 = 0, 0
        for num in nums:
            a0 = (a0 ^ num) & (~a1)
            a1 = (a1 ^ num) & (~a0)
        return a0

Single Number III

这里利用了一个找最低bit位的方法

两遍pass，第一遍找到两个结果的异或和，再找到这个异或和的最低位1，这个位就是两个结果不同的位，拿这个最低位1当bitmask去找就能分别找到两个数的结果了

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

        bitMask = sumOfTwoElements & (sumOfTwoElements ^ (sumOfTwoElements - 1))
        tmp = 0
        for num in nums:
            if num & bitMask:
                tmp ^= num

        result = [tmp, sumOfTwoElements ^ tmp]
        return result