Q1486 XOR Operation in an Array-简单-位操作

XOR Operation in an Array

Question

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.

Constraints:

• 1 <= n <= 1000
• 0 <= start <= 1000
• n == nums.length

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

Approach 1: 位操作

class Solution(object):
    def xorOperation(self, n: int, start: int) -> int:
        nums=0
        for i in range(n):
            nums ^= (start+2*i)
        return nums

def main():
    n=4
    start=3
    solution=Solution().xorOperation(n,start)
    print(solution)

if __name__ == "__main__":
    main()

复杂度分析

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