Question

961. N-Repeated Element in Size 2N Array

中文文档

Description

You are given an integer array nums with the following properties:

• nums.length == 2 * n.
• nums contains n + 1 unique elements.
• Exactly one element of nums is repeated n times.

Return _the element that is repeated _n_ times_.

 

Example 1:

Input: nums = [1,2,3,3]
Output: 3

Example 2:

Input: nums = [2,1,2,5,3,2]
Output: 2

Example 3:

Input: nums = [5,1,5,2,5,3,5,4]
Output: 5

 

Constraints:

• 2 <= n <= 5000
• nums.length == 2 * n
• 0 <= nums[i] <= 104
• nums contains n + 1 unique elements and one of them is repeated exactly n times.

Solutions

Solution 1

class Solution:
  def repeatedNTimes(self, nums: List[int]) -> int:
    s = set()
    for x in nums:
      if x in s:
        return x
      s.add(x)

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];
      }
    }
  }
}

class Solution {
public:
  int repeatedNTimes(vector<int>& nums) {
    unordered_set<int> s;
    for (int i = 0;; ++i) {
      if (s.count(nums[i])) {
        return nums[i];
      }
      s.insert(nums[i]);
    }
  }
};

func repeatedNTimes(nums []int) int {
  s := map[int]bool{}
  for i := 0; ; i++ {
    if s[nums[i]] {
      return nums[i]
    }
    s[nums[i]] = true
  }
}

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);
  }
}

/**
 * @param {number[]} nums
 * @return {number}
 */
var repeatedNTimes = function (nums) {
  const s = new Set();
  for (const x of nums) {
    if (s.has(x)) {
      return x;
    }
    s.add(x);
  }
};