[leetcode]421. Maximum XOR of Two Numbers in an Array

421. 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: 28Explanation: The maximum result is 5 ^ 25 = 28.

思路一:

利用a ^ b = c，而a ^ b ^ b = a, 则 c ^ b = a.

    int findMaximumXOR(vector<int>& nums) {        if( nums.size() < 2 ) return 0;        int maxNum = 0;        int flag = 0;                for( int i = 31; i >= 0; --i )        {//<span style="font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 13px; line-height: 19.5px;">从最大值maxNum最高位到最低位开始确定</span>            set<int> hash;                        flag |= (1<<i);            for( int x: nums )            {                hash.insert( flag & x );            }                        int tmp = maxNum | ( 1<<i );            for( int x: hash )            {                if( hash.find( x ^ tmp ) != hash.end() )                {                    maxNum = tmp;                    break;                }            }        }        return maxNum;    }

思路二:

Tire树,利用nums中的数构建tire树,然后依次查找数组nums中数的异或最大值

    struct Node{        Node * next[2];        Node()        {            next[0] = nullptr;            next[1] = nullptr;        }    };         void buildTireTree(Node* root, int x)    {        for( int i = 31; i >= 0; --i )        {            int flag =  ( x & (1<<i) ) ? 1 : 0;            if( root->next[flag] == nullptr )            {                root->next[flag] = new Node();            }            root = root->next[flag];        }    }        int findMaxXorInTire(Node* root, int x)    {        int result = 0;                for( int i = 31; i >= 0; --i )        {             int flag = ( x & ( 1<<i) )? 0: 1;            if( root->next[flag] != nullptr )            {                result |= (1<<i);                root = root->next[flag];            }            else                root = root->next[1-flag];        }        return result;    }    public:    int findMaximumXOR(vector<int>& nums) {       if( nums.size() < 2 ) return 0;       Node head;       for( int x: nums )       {           buildTireTree( &head, x );       }              int maxNum = 0;//INT_MIN;       for( int x: nums )       {           int m = findMaxXorInTire( &head, x );           maxNum = max( maxNum, m );       }       return maxNum;           }