Given an integer n
and an integer start
.
Define an array nums
where nums[i] = start + 2*i
(0-indexed) and n == nums.length
.
Return the bitwise XOR of all elements of nums
.
Example 1:
Input: n = 5, start = 0 Output: 8 Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8. Where "^" corresponds to bitwise XOR operator.
Example 2:
Input: n = 4, start = 3 Output: 8 Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.
Example 3:
Input: n = 1, start = 7 Output: 7
Example 4:
Input: n = 10, start = 5 Output: 2
Constraints:
1 <= n <= 1000
0 <= start <= 1000
n == nums.length
Related Topics:
Array, Bit Manipulation
// OJ: https://leetcode.com/problems/xor-operation-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int xorOperation(int n, int start) {
int ans = 0;
for (int i = 0; i < n; ++i) ans ^= start + 2 * i;
return ans;
}
};