Wins the offer - to find an array of number two appears only once

Title Description

In addition to an array of integers in two numbers, the other numbers appear twice. Please write a program to find these two figures appear only.

Thinking

The question I was going to use violence to solve, get a dictionary, then counts each number.

 

 

optimization

Space complexity is O (1)

With XOR, the same number of XOR is XOR 0,0 any number of any number. 

So that all or a number of different results for the two numbers appear only once

Then according to the XOR result, the number of bits to identify two different one (just need to find a like)

According to this bit packets, a group is 0, 1, as a group. In this way the two figures must have been assigned to different groups. The inside of each set of numbers, only a few appear only once, and the remaining number appeared twice ..... needless to say, were XOR

class Solution:
    def FindNumsAppearOnce(self, array):
        if len(array)<2:
            return False
        num = array[0] ^ array[1]
        for i in range(2,len(array)):
            num ^= array[i]
        
        # 找到二进制位不同的一位，只需要找出来一位就行
        i = 1
        while num%2 == 0:
            num = int(num//2)
            i += 1
        
        # 按照上面找到的二进制位分组异或
        num1 = 0
        num2 = 0
        for num in array:
            if num%(2**i) == (2**(i-1)):
                num1 ^= num
            else:
                num2 ^= num
        return num1,num2