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:
4 <= A.length <= 100000 <= A[i] < 10000A.lengthis even
题目描述：求一个长度为 2N 数组中重复 N 次的元素值
题目分析：很简单，把每一个出现过的不同元素进行统计，重复次数大于等于 2 的元素即为我们所求。
python 代码：
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++ 代码：
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]; } } } };
