1486. XOR Operation in an Array

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

---

给定start和n，将所有的start+2i进行异或。简单题，直接照做：

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;
	}
};

本代码提交AC，用时0MS。