题目如下:
Given an integer
n
and an integerstart
.Define an array
nums
wherenums[i] = start + 2*i
(0-indexed) andn == 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: 7Example 4:
Input: n = 10, start = 5 Output: 2Constraints:
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