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69 lines
1.3 KiB
Markdown
69 lines
1.3 KiB
Markdown
# [1486. XOR Operation in an Array](https://leetcode.com/problems/xor-operation-in-an-array/)
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## 题目
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Given an integer `n` and an integer `start`.
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Define an array `nums` where `nums[i] = start + 2*i` (0-indexed) and `n == nums.length`.
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Return the bitwise XOR of all elements of `nums`.
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**Example 1:**
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```
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Input: n = 5, start = 0
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Output: 8
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Explanation:Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
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Where "^" corresponds to bitwise XOR operator.
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```
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**Example 2:**
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```
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Input: n = 4, start = 3
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Output: 8
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Explanation:Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.
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```
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**Example 3:**
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```
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Input: n = 1, start = 7
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Output: 7
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```
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**Example 4:**
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```
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Input: n = 10, start = 5
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Output: 2
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```
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**Constraints:**
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- `1 <= n <= 1000`
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- `0 <= start <= 1000`
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- `n == nums.length`
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## 题目大意
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给你两个整数，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（XOR）后得到的结果。
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## 解题思路
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- 简单题。按照题意，一层循环依次累积异或数组中每个元素。
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## 代码
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```go
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package leetcode
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func xorOperation(n int, start int) int {
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res := 0
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for i := 0; i < n; i++ {
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res ^= start + 2*i
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}
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return res
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}
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``` |