N-Repeated Element in Size 2N Array

// Accepted solution for LeetCode #961: N-Repeated Element in Size 2N Array
class Solution {
    public int repeatedNTimes(int[] nums) {
        Set<Integer> s = new HashSet<>(nums.length / 2 + 1);
        for (int i = 0;; ++i) {
            if (!s.add(nums[i])) {
                return nums[i];
            }
        }
    }
}

// Accepted solution for LeetCode #961: N-Repeated Element in Size 2N Array
func repeatedNTimes(nums []int) int {
	s := map[int]bool{}
	for i := 0; ; i++ {
		if s[nums[i]] {
			return nums[i]
		}
		s[nums[i]] = true
	}
}

# Accepted solution for LeetCode #961: N-Repeated Element in Size 2N Array
class Solution:
    def repeatedNTimes(self, nums: List[int]) -> int:
        s = set()
        for x in nums:
            if x in s:
                return x
            s.add(x)

// Accepted solution for LeetCode #961: N-Repeated Element in Size 2N Array
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn repeated_n_times(a: Vec<i32>) -> i32 {
        let mut hs: HashSet<i32> = HashSet::new();
        for x in a {
            if !hs.insert(x) {
                return x;
            }
        }
        unreachable!()
    }
}

#[test]
fn test() {
    let a = vec![1, 2, 3, 3];
    assert_eq!(Solution::repeated_n_times(a), 3);
    let a = vec![2, 1, 2, 5, 3, 2];
    assert_eq!(Solution::repeated_n_times(a), 2);
    let a = vec![5, 1, 5, 2, 5, 3, 5, 4];
    assert_eq!(Solution::repeated_n_times(a), 5);
}

// Accepted solution for LeetCode #961: N-Repeated Element in Size 2N Array
function repeatedNTimes(nums: number[]): number {
    const s: Set<number> = new Set();
    for (const x of nums) {
        if (s.has(x)) {
            return x;
        }
        s.add(x);
    }
}