XOR Operation in an Array
You are 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.
Constraints:
1 <= n <= 1000
0 <= start <= 1000
n == nums.length
思路一：这题有点奇怪，直接教你怎么做...
public int xorOperation(int n, int start) {
int result = 0;
for (int i = 0; i < n; i++) {
result ^= start + 2 * i;
}
return result;
}
思路二：看了题解，发现从数学上有 O(1) 优化的优化解
标签：XOR,nums,int,start,1486,easy,Array,where,leetcode From： https://www.cnblogs.com/iyiluo/p/16884594.html