[Swift]LeetCode421. 数组中两个数的最大异或值 | Maximum XOR of Two Numbers in an Array

原文地址：https://www.cnblogs.com/strengthen/p/10333729.html 

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 ≤ i, j < 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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给定一个非空数组，数组中元素为 a0, a1, a2, … , an-1，其中 0 ≤ ai < 231 。

找到 ai 和aj 最大的异或 (XOR) 运算结果，其中0 ≤ i,  j < n 。

你能在O(n)的时间解决这个问题吗？

示例:

输入: [3, 10, 5, 25, 2, 8]

输出: 28

解释: 最大的结果是 5 ^ 25 = 28.

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196ms

class Solution {
    func findMaximumXOR(_ nums: [Int]) -> Int {
        var res:Int = 0
        var mask:Int = 0
        for i in (0...31).reversed()
        {
            mask |= (1 << i)
            var s:Set<Int> = Set<Int>()
            for num in nums
            {
                s.insert(num & mask)
            }
            var t:Int = res | (1 << i)
            for prefix in s
            {
                if s.contains(t ^ prefix)
                {
                    res = t
                    break
                }
            }
        }
        return res
    }
}

---

872ms

class TrieNode {
    var children = [Int: TrieNode]()
    var end: Bool
    var value: Int = -1
    
    init() {
        end = false
    }
    
    func hasChild(_ key: Int) -> Bool {
        return children[key] != nil
    }
    
    func makeChild(_ key: Int) {
        children[key] = TrieNode()
    }
    
    func getChild(_ key: Int) -> TrieNode {
        return children[key]!
    }
}


class Solution {
    
    let INT_SIZE = 32
    var root = TrieNode()
    
    func findMaximumXOR(_ nums: [Int]) -> Int {
        
        var ans = 0
        insert(nums[0])
        
        for i in 1..<nums.count {
            ans = max(ans, query(nums[i]))
            insert(nums[i])
        }
        
        return ans
        
    }
    
    func query(_ key: Int) -> Int {
        var current = root 
        
        for i in (0..<INT_SIZE).reversed() {
            let current_bit = (key & 1<<i) >= 1 ? 1 : 0
            let next: TrieNode?
            if current.hasChild(1-current_bit) {
                next = current.getChild(1-current_bit)
            } else {
                next = current.getChild(current_bit)
            } 
            
            current = next!
        }
        
        return key ^ current.value
        
    }
    
    func insert(_ num: Int) {
        
        var current = root
        
        for i in (0..<INT_SIZE).reversed() {
            let key = (num & 1<<i) >= 1 ? 1 : 0
            if !current.hasChild(key) {
                current.makeChild(key)
            }
            
            current = current.getChild(key)
        }
        
        current.end = true
        current.value = num
    }
}