In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.

Example 1:

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

Example 2:

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

Example 3:

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

Note:

1. 4 <= A.length <= 10000
2. 0 <= A[i] < 10000
3. A.length is even

題目描述：求一個長度為 2N 陣列中重複 N 次的元素值

題目分析：很簡單，把每一個出現過的不同元素進行統計，重複次數大於等於 2 的元素即為我們所求。

python 程式碼：

class Solution(object):
    def repeatedNTimes(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        A_length = len(A)
        for i in range(A_length):
            if A.count(A[i]) >= 2:
                return A[i]
            else:
                continue
 

C++ 程式碼：

class Solution {
public:
    int repeatedNTimes(vector<int>& A) {
        for(int i = 0; i < A.size(); i++){
            if(count(A.begin(),A.end(),A[i]) >= 2){
                return A[i];
            }
        }
    }
};