260. Single Number III

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

Follow up: Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

 

Example 1:

Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation:  [5, 3] is also a valid answer.

Example 2:

Input: nums = [-1,0]
Output: [-1,0]

Example 3:

Input: nums = [0,1]
Output: [1,0]

 

Constraints:

  • 2 <= nums.length <= 3 * 104
  • -231 <= nums[i] <= 231 - 1
  • Each integer in nums will appear twice, only two integers will appear once.

Rust Solution

struct Solution;

impl Solution {
    fn single_number(nums: Vec<i32>) -> Vec<i32> {
        let mut bitmask = 0;
        for &num in &nums {
            bitmask ^= num;
        }
        let diff = bitmask & (-bitmask);
        let mut x = 0;
        for &num in &nums {
            if diff & num != 0 {
                x ^= num;
            }
        }
        vec![x, bitmask ^ x]
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 1, 3, 2, 5];
    let res = vec![3, 5];
    assert_eq!(Solution::single_number(nums), res);
}

Having problems with this solution? Click here to submit an issue on github.