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：
思路：
從最高位開始，統(tǒng)計nums各元素在此位的值是1是0宏娄，依次mask從10000, 11000, 11100 ...
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
public class Solution {
public int findMaximumXOR(int[] nums) {
int max = 0, mask = 0;
for (int i = 31; i >= 0; i--) {
mask |= (1 << i); // ex, mask: 10000 -> 11000 -> 11100 -> 11110 -> 11111
HashSet<Integer> set = new HashSet<Integer>();
for (int num : nums) {
set.add(num & mask); // reserve Left bits and ignore Right bits
}
/* Use 0 to keep the bit, 1 to find XOR
* 0 ^ 0 = 0
* 0 ^ 1 = 1
* 1 ^ 0 = 1
* 1 ^ 1 = 0
*/
int tmp = max | (1 << i); // 假裝能達到最大1，左邊的位從max結(jié)果保留
// in each iteration, there are pair(s) whose Left bits can XOR to max
for (int prefix : set) {
if (set.contains(tmp ^ prefix)) { // a ^ b = c, b ^ c = a, c ^ a = b, 看set(候選輸入)有沒有這樣的輸入能達到tmp
max = tmp;
}
}
}
return max;
}
}
