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  • 【leetcode】1486. XOR Operation in an Array

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