421. Maximum XOR of Two Numbers in an Array

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本文介绍了一种利用前缀树解决寻找数组中两个数最大异或值的方法。通过构建前缀树来存储数组中的每个元素，并通过比较每个元素与树中其他元素找到最大的异或结果。

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Maximum XOR of Two Numbers in an Array
- description
- implementation

421. Maximum XOR of Two Numbers in an Array

description

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.

Use prextree to finish this task.

1 build prefix-tree which contain every element in vector from significant bit to least bit which is binary tree.

2 compare every element in vector to binary tree to find the largest element. Here is trick is that the maximum is with significant different number.

3 go throught the whole vector to get the maximum result.

implementation

class TrieNode {
public:
    TrieNode* next[2];
    TrieNode() {
        this->next[0] = NULL, this->next[1] = NULL;
    }
};

class Solution {
public:
    TrieNode* root = NULL;

    void buildTrieTree(vector<int>& nums) {
        int len = nums.size();
        if(!len) return;
        this->root = new TrieNode();
        TrieNode* cur = this->root;
        for(int ind = 0; ind < len; ind++) {
            int tmp = nums[ind];
            cur = this->root;
            for(int k = 31; k >= 0; k--) {
                int tree_node = (tmp >> k) & 1;
                if(!cur->next[tree_node]) 
                    cur->next[tree_node] = new TrieNode();
                cur = cur->next[tree_node];    
            }
        }
    }

    int calculateMax(int element) {
        int res = 0;
        TrieNode* cur = this->root;
        for(int i = 31; i >= 0; i--) {
            int dir = 1 - (element >> i) & 1; 
            if(cur->next[dir]) {
                res <<= 1;
                res |= 1;
            }   
            else {
                res <<= 1;
                dir = 1 - dir;
            }
            cur = cur->next[dir];
        }
        return res;
    }

    int findMaximumXOR(vector<int>& nums) {
        buildTrieTree(nums);
        int res = 0;
        for(auto ele:nums)
            res = max(calculateMax(ele), res);
        return res;    
    }
};