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

在大小為 2N 的數組 A 中有 N+1 個不同的元素，其中有一個元素重復了 N 次。

返回重復了 N 次的那個元素。

示例 1：

輸入：[1,2,3,3]
輸出：3

示例 2：

輸入：[2,1,2,5,3,2]
輸出：2

示例 3：

輸入：[5,1,5,2,5,3,5,4]
輸出：5

提示：

1. 4 <= A.length <= 10000
2. 0 <= A[i] < 10000
3. A.length 為偶數

 1
 
 class Solution {
 2     func repeatedNTimes(_ A: [Int]) -> Int {
 3         var set:Set<Int> = Set<Int>()
 4         for x in A
 5         {
 6             if set.contains(x)
 7             {
 8                 return x
 9             }
10             set.insert(x)
11         }
 
12         return -1
13     }
14 }

300ms

 1 class Solution {
 2     func repeatedNTimes(_ A: [Int]) -> Int {
 3         var n:Int = A.count
 4         var f:[Int] = [Int](repeating:0,count:100000)
 5         for v in A
 6         {
 7             f[v] += 1
 8         }
 9         
10         for i in 0..<100000
11         {
12             if f[i] > 1
13             {
14                 return i
15             }
16         }
17         return -1
18     }
19 }

[Swift Weekly Contest 116]LeetCode961. 重復 N 次的元素 | N-Repeated Element in Size 2N Array