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

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.

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.

196ms

 1 class Solution {
 2     func findMaximumXOR(_ nums: [Int]) -> Int {
 3         var res:Int = 0
 4         var mask:Int = 0
 5         for i in (0...31).reversed()
 6         {
 7             mask |= (1 << i)
 8             var s:Set<Int> = Set<Int>()
 9             for num in nums
10             {
11                 s.insert(num & mask)
12             }
13             var t:Int = res | (1 << i)
14             for prefix in s
15             {
16                 if s.contains(t ^ prefix)
17                 {
18                     res = t
19                     break
20                 }
21             }
22         }
23         return res
24     }
25 }

 1 class TrieNode {
 2     var children = [Int: TrieNode]()
 3     var end: Bool
 4     var value: Int = -1
 5
 6     init() {
 7         end = false
 8     }
 9
10     func hasChild(_ key: Int) -> Bool {
11         return children[key] != nil
12     }
13
14     func makeChild(_ key: Int) {
15         children[key] = TrieNode()
16     }
17
18     func getChild(_ key: Int) -> TrieNode {
19         return children[key]!
20     }
21 }
22
23
24 class Solution {
25
26     let INT_SIZE = 32
27     var root = TrieNode()
28
29     func findMaximumXOR(_ nums: [Int]) -> Int {
30
31         var ans = 0
32         insert(nums[0])
33
34         for i in 1..<nums.count {
35             ans = max(ans, query(nums[i]))
36             insert(nums[i])
37         }
38
39         return ans
40
41     }
42
43     func query(_ key: Int) -> Int {
44         var current = root
45
46         for i in (0..<INT_SIZE).reversed() {
47             let current_bit = (key & 1<<i) >= 1 ? 1 : 0
48             let next: TrieNode?
49             if current.hasChild(1-current_bit) {
50                 next = current.getChild(1-current_bit)
51             } else {
52                 next = current.getChild(current_bit)
53             }
54
55             current = next!
56         }
57
58         return key ^ current.value
59
60     }
61
62     func insert(_ num: Int) {
63
64         var current = root
65
66         for i in (0..<INT_SIZE).reversed() {
67             let key = (num & 1<<i) >= 1 ? 1 : 0
68             if !current.hasChild(key) {
69                 current.makeChild(key)
70             }
71
72             current = current.getChild(key)
73         }
74
75         current.end = true
76         current.value = num
77     }
78 }

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