【leetcode】1486. XOR Operation in an Array

标签：ret   turn   res   span   bsp   constrain   integer   length   The   

题目如下：

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

解题思路：很简单的题目，依次做XOR操作就好了。

代码如下：

class Solution(object):
    def xorOperation(self, n, start):
        """
        :type n: int
        :type start: int
        :rtype: int
        """
        res = start
        i = 1
        while i < n:
            res ^= (start + 2*i)
            i += 1
        return res

【leetcode】1486. XOR Operation in an Array

标签：ret   turn   res   span   bsp   constrain   integer   length   The   

原文地址：https://www.cnblogs.com/seyjs/p/13176660.html