[LeetCode]421. Maximum XOR of Two Numbers in an Array

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https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/?tab=Description

找出数组中两个数异或的最大值

解法一：

Trie树

字典树两个函数：1、search；2、insert

注意记录isNum的地方要用“或”，避免覆盖

public class Solution {    public int findMaximumXOR(int[] nums) {        TreeNode root = new TreeNode();        int res = Integer.MIN_VALUE;        for (int num : nums) {            res = Math.max(res, search(root, num, 31));            insert(root, num, 31);        }        return res;    }    private int search(TreeNode root, int num, int index) {        if (index < 0 || root == null) {            return 0;        }        boolean zero = (num & (1 << index)) == 0;        if (zero && root.one != null) {            return (1 << index) + search(root.one, num, index - 1);        } else if (!zero && root.zero != null) {            return (1 << index) + search(root.zero, num, index - 1);        } else {            return Math.max(search(root.one, num, index - 1), search(root.zero, num, index - 1));        }    }    private void insert(TreeNode root, int num, int index) {        if (index < 0) {            return;        }        boolean zero = (num & (1 << index)) == 0;        if (zero && root.zero == null) {            root.zero = new TreeNode();            root.zero.isNum |= (index == 0);        } else if (!zero && root.one == null) {            root.one = new TreeNode();            root.one.isNum |= (index == 0);        }        insert(zero ? root.zero : root.one, num, index - 1);    }}class TreeNode {    boolean isNum;    TreeNode zero;    TreeNode one;}

解法二：

位运算

外层遍历找到最高位到第i位异或的最大值，greedyTry是用来假设当前位异或之后可以取一，如果确实当前位异或之后取一的话，那么set中两个数异或值为greedyTry，借助注释性质可得解

public class Solution {    public int findMaximumXOR(int[] nums) {        int res = 0;        int mask = 0;        for (int i = 31; i >= 0; i--) {            HashSet<Integer> set = new HashSet();            mask |= (1 << i);            for (int num : nums) {                set.add((num & mask));            }            int greedyTry = (res | (1 << i));            for (int leftNum : set) {                // a ^ b = c ==>> a ^ c = b                if (set.contains((leftNum ^ greedyTry))) {                    res = greedyTry;                    break;                }            }        }        return res;    }}

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