C++ 1486. XOR Operation in an Array【Array/Bit Manipulation】简单

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

题意：两个整数 n 和 start 。数组 nums 定义为：nums[i] = start + 2 * i（下标从 0 开始）且 n == nums.length 。请返回 nums 中所有元素按位异或后得到的结果。

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解法 常量空间

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;
    }
};

执行效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户
内存消耗：6.3 MB, 在所有 C++ 提交中击败了7.13% 的用户

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