SIngle number in a list

Given an array of integers, every element appearstwiceexcept 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
        """
        return reduce(lambda x, y : x ^ y, nums)

Other ways:

class Solution(object):
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #////////////////////
        # using extra memory
        #////////////////////
        '''
        storeH = {}
        for item in nums:
            # keep a count of the frequency of each element
            if item in storeH:
                storeH[item] += 1
            else:
                storeH[item] = 1

        # Now traverse the hash
        for (k, v) in storeH.iteritems():
            #print k, ' -> ', v
            if v == 1:
                return k
        '''
        #////////////////////////////////////
        # without using xtra memory
        # use XOR. XOR is also commutative
        #////////////////////////////////////
        result = 0
        for item in nums:
            result ^= item
        return result