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.
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
EXPLANATION:
这道题目就比较简单了. 直接根据题意, 然后按照对应的方法写出来即可. 先把数组列出来, 然后再把每一位进行异或就行.
SOLUTION:
class Solution {
fun xorOperation(n: Int, start: Int): Int {
if (n ==0) return 0
var array : IntArray = IntArray(n)
for (i in array.indices) {
array[i] = start + i * 2
}
var result = array[0];
for (i in 1..array.size-1) {
result = result xor array[i]
}
return result
}
}