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.
Solution 1: Bit Manipulation:
这题真心有点难,get the max XOR bit by bit
note that if A^B=C, then A^C=B, B^C=A
public class Solution {
public int findMaximumXOR(int[] nums) {
int mask = 0, max = 0;
for (int i=31; i>=0; i--) {
mask |= 1<<i;
HashSet<Integer> prefixes = new HashSet<Integer>();
for (int each : nums) {
prefixes.add(each & mask); // reserve Left bits and ignore Right bits
}
int tmpMax = max | (1<<i);
//possible new max, for example: max right now is 11000, then tmpMax=11100, max is 11100, tmpMax is 11110
for (int prefix : prefixes) {
if (prefixes.contains(tmpMax^prefix))
max = tmpMax;
}
}
return max;
}
}
Solution 2: Trie, (未研究)
https://discuss.leetcode.com/topic/63207/java-o-n-solution-using-trie
https://discuss.leetcode.com/topic/64753/31ms-o-n-java-solution-using-trie