421. Maximum XOR of Two Numbers in an Array
2016-12-04 16:25
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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:
分析:采用trie,字典树,即将这些数 根据其bit位依次从最高位开始存储。root不存储任何元素。
一边存储的时候一边计算其与已经存储在字典树中的值得最大值。
public class Solution {
class Trie{
Trie[] child=new Trie[2];
}
public int findMaximumXOR(int[] nums) {
int n=nums.length;
//construct trie tree
Trie root=new Trie();
int temp;
int max=Integer.MIN_VALUE;
for(int num:nums){
temp=0;
Trie current=root;
Trie num_trie=root;
for(int i=31;i>=0;i--){
int flag=(num>>>i)&1;
if(current.child[flag]==null)
current.child[flag]=new Trie();
current=current.child[flag];
if(num_trie.child[1^flag]!=null){
temp+=(1<<i);
num_trie=num_trie.child[1^flag];
}else{
num_trie=num_trie.child[flag];
}
}
max=max>temp?max:temp;
}
return max;
}
}
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.
分析:采用trie,字典树,即将这些数 根据其bit位依次从最高位开始存储。root不存储任何元素。
一边存储的时候一边计算其与已经存储在字典树中的值得最大值。
public class Solution {
class Trie{
Trie[] child=new Trie[2];
}
public int findMaximumXOR(int[] nums) {
int n=nums.length;
//construct trie tree
Trie root=new Trie();
int temp;
int max=Integer.MIN_VALUE;
for(int num:nums){
temp=0;
Trie current=root;
Trie num_trie=root;
for(int i=31;i>=0;i--){
int flag=(num>>>i)&1;
if(current.child[flag]==null)
current.child[flag]=new Trie();
current=current.child[flag];
if(num_trie.child[1^flag]!=null){
temp+=(1<<i);
num_trie=num_trie.child[1^flag];
}else{
num_trie=num_trie.child[flag];
}
}
max=max>temp?max:temp;
}
return max;
}
}
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