Description

Given a non-empty array of numbers, a₀, a₁, a₂, … , a_n-1, where 0 ≤ a_i < 2³¹.

Find the maximum result of a_i XOR a_j, 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.

Solution

Trie, time O(n), space O(n)

题意还是很明白的，找到两个数，二进制表达差的越多越好（大概这个意思）。但是怎么达到O(n)复杂度的确是个问题。遍历数组，对于每个i，找到和它差的最多的j？但这样就是O(n ^ 2)的复杂度啊。仔细想来可以用Trie来表示整个array，然后对于每一个i，只需要常数时间就能找到对应的j了。

注意要边搜索边累加XOR，否则到最后在计算i ^ j的话会TLE……

class Solution {
    public int findMaximumXOR(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }
        
        // build Trie
        TrieNode root = new TrieNode();
        for (int n : nums) {
            TrieNode pre = root;
            
            for (int i = 31; i >= 0; --i) {
                int bit = (n >> i) & 1;
                if (pre.children[bit] == null) {
                    pre.children[bit] = new TrieNode();
                }
                pre = pre.children[bit];
            }
        }
        // search Trie
        int maxXOR = Integer.MIN_VALUE;
        
        for (int n : nums) {
            TrieNode pre = root;
            int xor = 0;
            
            for (int i = 31; i >= 0; --i) {
                int bit = (n >> i) & 1;
                if (pre.children[bit ^ 1] != null) {
                    pre = pre.children[bit ^ 1];
                    xor += 1 << i;  // add to xor
                } else {
                    pre = pre.children[bit];
                }
            }
            
            maxXOR = Math.max(maxXOR, xor);
        }
        
        return maxXOR;
    }
    
    class TrieNode {
        TrieNode[] children;
        
        public TrieNode() {
            children = new TrieNode[2];
        }
    }
}