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: 28Explanation: The maximum result is 5 ^ 25 = 28.

思路：按照位来算

/** 异或出来的结果就是31！种* 实际上我们可以一个bit一个bit来，这样可以减少一点重复计算* 比如11....是不行的，那么11..1..就不要算了*/
public class Solution {public int findMaximumXOR(int[] nums) {int prefix = 0, max = 0;for(int i=31; i>=0; i--) {Set<Integer> set = new HashSet<Integer>();prefix = prefix | (1<<i);for(int num : nums)set.add(num & prefix);// 计算可能的更大值，基于以下定理：a^b^b=a，// 如果set里面有2个数异或可以得到possibleMax，那么possibleMax和set里的一个数异或得到的就是set里面的另外一个数int possibleMax = max | (1<<i);for(int num : set)if(set.contains(possibleMax ^ num)) {max = possibleMax;break;}}return max;}
}

Trie树版本

/** 需要两两比较，又是按照一位一位来计算的，所以考虑Trie* 总体思路还是按位来算*/
public class trie1 {class Trie {Trie[] children = new Trie[2];}public int findMaximumXOR(int[] nums) {// build up TrieTrie root = new Trie();for(int num : nums) {Trie cur = root;for(int i=31; i>=0; i--) {int bit = (num >> i) & i;if(cur.children[bit] == null)cur.children[bit] = new Trie();cur = cur.children[bit];}}// find bit by bitint max = Integer.MIN_VALUE;for(int num : nums) {Trie cur = root;int sum = 0;for(int i=31; i>=0; i--) {int bit = (num >> i) & i;int xor = bit ^ 1;if(cur.children[xor] != null) {cur = cur.children[xor];sum += (1 << i);} else if(cur.children[bit] != null)cur = cur.children[bit];}max = Math.max(max, sum);}return max;}
}

用Object数组表示的Trie树版本

/** 新建Class Trie会超时*/
public class Solution {public int findMaximumXOR(int[] nums) {// build up TrieObject[] root = {null, null};for(int num : nums) {Object[] cur = root;for(int i=31; i>=0; i--) {int bit = (num >> i) & 1;if(cur[bit] == null)cur[bit] = new Object[]{null, null};cur = (Object[]) cur[bit];}}// find bit by bitint max = Integer.MIN_VALUE;for(int num : nums) {Object[] cur = root;int sum = 0;for(int i=31; i>=0; i--) {int bit = (num >> i) & 1;int xor = bit ^ 1;if(cur[xor] != null) {cur = (Object[]) cur[xor];sum += (1 << i);} else if(cur[bit] != null)cur = (Object[]) cur[bit];}max = Math.max(max, sum);}return max;}
}

2刷：一般这样的bit运算都有个套路就是要按位来求，反正int就32位，

求到第i位时判断有没有2个数的前缀异或得到当前的最大值，想到了前缀，Trie树就自然而然想到了

/** 按位来求* 求到第i位时判断有没有2个数的前缀异或得到当前的最大值*/
public class Solution {public int findMaximumXOR(int[] nums) {int max = 0, prefix = 0;for(int i=31; i>=0; i--) {// 把前缀放到set里面prefix = prefix | (1<<i);Set<Integer> s = new HashSet<Integer>();for(int num : nums)        s.add(num & prefix);int tmp = max | (1<<i);for(int num : s)if(s.contains(tmp ^ num)) {   // 有2个前缀可以异或成当前可能的最大值max = tmp;break;}}return max;}
}

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