首先，我們想一個(gè)簡(jiǎn)單的問(wèn)題。
Given an array A, find the two numbers a, b in A that makes a * b == 1223.
我們可能首先想這么寫(xiě)

for i in range(len(nums)):
   for j in range(i + 1, len(nums)):
       if nums[i] * nums[j] == 1223:
           return True

但是我們想一下，其實(shí)可以?xún)?yōu)化位O(n), 用一個(gè)set就行了

set_ = set(nums)
for num in nums:
    if 1223 // num in set_:
            return True

然后我們看這個(gè)問(wèn)題
https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/#/solutions
實(shí)際上就是把上面的過(guò)程重復(fù)了32遍

class Solution(object):
    def findMaximumXOR(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        max = 0
        for i in range(31, -1, -1):
            max <<= 1
            prefixes = {num >> i for num in nums}
            expect_max = max | 1
            for prefix in prefixes: # in prefixes, we want to find the a ^ b == expect_max. 
                if expect_max ^ prefix in prefixes:
                    max = expect_max
        return max