Leetcode Maximum XOR of Two Numbers in an Array

本文介绍了一种在O(n)时间复杂度内找到数组中两个数的最大XOR值的方法。通过逐位遍历并利用哈希集合判断是否存在能使当前前缀达到最大值的数,最终确定最大XOR值。

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Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231.

Find the maximum result of ai XOR aj, where 0 ≤ ij < n.

Could you do this in O(n) runtime?

Example:

Input: [3, 10, 5, 25, 2, 8]

Output: 28

Explanation: The maximum result is 5 ^ 25 = 28.

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Cycle bit by bit. For every bit, we could know the largest prefix expected. If the expected largest prefix could be xor by some two nums (implemented by hashSet), we could know the final answer. The complexity is O(n)

class Solution:
    def findMaximumXOR(self, nums):
        pre, mask = 0, 0
        for i in range(31, -1, -1):
            one = (1 << i)
            mask |= one
            exist = set()
            cur_largest = pre | one

            for num in nums:
                cur = num & mask
                if ((cur ^ cur_largest) in exist):
                    pre = cur_largest
                else:
                    exist.add(cur)
        return pre
s = Solution()
print(s.findMaximumXOR([3, 10, 5, 25, 2, 8]))

 

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