題目描述
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
題目思路
- 思路一壶笼、考察整數(shù)的異或運算
class Solution {
public:
int xorOperation(int n, int start) {
int res = start;
int tmp = 0;
for(int i = 1; i < n; ++i){
tmp = start + 2 * i;
res = res ^ tmp;
}
return res;
}
};
