[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: 28

Explanation: 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;

}