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package class06 ;
// 测试链接 : https://leetcode.com/problems/maximum-xor-with-an-element-from-array/
public class Code03_MaximumXorWithAnElementFromArray {
public static int [ ] maximizeXor ( int [ ] nums , int [ ] [ ] queries ) {
int N = nums . length ;
NumTrie trie = new NumTrie ( ) ;
for ( int i = 0 ; i < N ; i + + ) {
trie . add ( nums [ i ] ) ;
}
int M = queries . length ;
int [ ] ans = new int [ M ] ;
for ( int i = 0 ; i < M ; i + + ) {
ans [ i ] = trie . maxXorWithXBehindM ( queries [ i ] [ 0 ] , queries [ i ] [ 1 ] ) ;
}
return ans ;
}
public static class Node {
public int min ;
public Node [ ] nexts ;
public Node ( ) {
min = Integer . MAX_VALUE ;
nexts = new Node [ 2 ] ;
}
}
public static class NumTrie {
public Node head = new Node ( ) ;
public void add ( int num ) {
Node cur = head ;
head . min = Math . min ( head . min , num ) ;
for ( int move = 30 ; move > = 0 ; move - - ) {
int path = ( ( num > > move ) & 1 ) ;
cur . nexts [ path ] = cur . nexts [ path ] = = null ? new Node ( ) : cur . nexts [ path ] ;
cur = cur . nexts [ path ] ;
cur . min = Math . min ( cur . min , num ) ;
}
}
// 这个结构中,已经收集了一票数字
// 请返回哪个数字与X异或的结果最大, 返回最大结果
// 但是,只有<=m的数字, 可以被考虑
public int maxXorWithXBehindM ( int x , int m ) {
if ( head . min > m ) {
return - 1 ;
}
// 一定存在某个数可以和x结合
Node cur = head ;
int ans = 0 ;
for ( int move = 30 ; move > = 0 ; move - - ) {
int path = ( x > > move ) & 1 ;
// 期待遇到的东西
int best = ( path ^ 1 ) ;
best ^ = ( cur . nexts [ best ] = = null | | cur . nexts [ best ] . min > m ) ? 1 : 0 ;
// best变成了实际遇到的
ans | = ( path ^ best ) < < move ;
cur = cur . nexts [ best ] ;
}
return ans ;
}
}
}