1486. X O R Operation in an Array

1486. XOR Operation in an Array #

题目 #

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

题目大意 #

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。

解题思路 #

  • 简单题。按照题意,一层循环依次累积异或数组中每个元素。

代码 #

package leetcode

func xorOperation(n int, start int) int {
	res := 0
	for i := 0; i < n; i++ {
		res ^= start + 2*i
	}
	return res
}

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