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.

这道题用单词搜索树思路比较清晰，存的是当前位的值，0和1。查找的时候，找能不能找到跟当前位异或出1的位，如果可以，temp加上那一位，最后判断跟max的大小。如果不加限制条件，会出现TEL的情况。代码如下：

public class Solution {
    class Trie {
        Trie[] child;
        public Trie() {
            child = new Trie[2];
        }
    }
    
    public int findMaximumXOR(int[] nums) {
        Trie root = new Trie();
        for (int num: nums) {
            Trie currTrie = root;
            for (int i = 31; i >= 0; i --) {
                int currBit = (num >>> i) & 1;
                if (currTrie.child[currBit] == null) {
                    currTrie.child[currBit] = new Trie();
                }
                currTrie = currTrie.child[currBit];
            }
        }
        int max = 0;
        for (int num: nums) {
            Trie currTrie = root;
            int temp = 0;
            for (int i = 31; i >= 0; i --) {
                int currBit = (num >>> i) & 1;
                if (currTrie.child[currBit ^ 1] != null) {
                    currTrie = currTrie.child[currBit ^ 1];
                    temp += (1 << i);
                } else {
                    currTrie = currTrie.child[currBit];
                }
                // for this case: even if all left bits results are 1s, curSum still cannot catch up max value
                if (temp < max && max - temp >= (1 << i) - 1) {
                    break;
                }
            }
            max = Math.max(max, temp);
        }
        return max;
    }
}