LeetCode 421. Maximum XOR of Two Numbers in an Array 解题报告（Trie树）

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.

思路：利用数组中每个元素二进制的表示形式建一棵前缀树。然后再对数组中的每一个元素，在前缀树中去搜索最大的异或值。在每次异或的时候，我们都沿着异或值最大的分支往下走，所以helper函数的时间复杂度是O(32) 。因此整个算法的时间复杂度就是O(32n) = O(n)

详解地址：https://www.acwing.com/solution/LeetCode/content/549/

class Solution {
public:
    class TreeNode
    {
    public:
        TreeNode* next[2];
        TreeNode()
        {
            next[0] = NULL;
            next[1] = NULL;
        }
    };
    int findMaximumXOR(vector& nums)
    {
        TreeNode* root = build(nums);
        int res = 0;
        for(int i = 0; i < nums.size(); i++)
            res = max(res, find(root, nums[i]));
        return res;
    }
    TreeNode* build(vector& nums)
    {
        TreeNode* root = new TreeNode(), *cur;
        for(int i = 0; i < nums.size(); i++)
        {
            cur = root;
            for(int j = 30; j >= 0; j--)
            {
                int k = (nums[i] >> j) & 1;
                if(cur->next[k] == NULL)
                    cur->next[k] = new TreeNode();
                cur = cur->next[k];
            }
        }
        return root;
    }
    int find(TreeNode* cur, int num)
    {
        int res = 0;
        for(int i = 30; i >= 0; i--)
        {
            int k = (num >> i) & 1;
            if(cur->next[!k])
            {
                res <<= 1;
                res |= 1;
                cur = cur->next[!k];
            }
            else
            {
                res <<= 1;
                cur = cur->next[k];
            }
        }
        return res;
    }

};