Easy - 1486. XOR Operation in an Array - 《Solve Leetcode Bit Manipulation Problems》

Solution

Difficulty: Easy

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

Solution

Language: Java

class Solution {
    // n based on 1
    public int xorOperation(int n, int start) {
        if (n == 1) return start;
        int num = start + 2 * (n - 1);
        return num ^ xorOperation(n - 1, start);
    }
}