README_EN.md

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

Python3

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

Java

class Solution {
    public int repeatedNTimes(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int num : nums) {
            if (s.contains(num)) {
                return num;
            }
            s.add(num);
        }
        return -1;
    }
}

C++

class Solution {
public:
    int repeatedNTimes(vector<int>& nums) {
        unordered_set<int> s;
        for (auto& num : nums) {
            if (s.find(num) != s.end()) {
                return num;
            }
            s.insert(num);
        }
        return -1;
    }
};

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var repeatedNTimes = function (nums) {
    const s = new Set();
    for (const num of nums) {
        if (s.has(num)) {
            return num;
        }
        s.add(num);
    }
    return -1;
};

...