#### QUESTION:

You are given a 0-indexed integer array nums. In one operation, you may do the following:

Choose two integers in nums that are equal. Remove both integers from nums, forming a pair. The operation is done on nums as many times as possible.

Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

**Example 1:**

```
Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.
```

**Example 2:**

```
Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.
```

**Example 3:**

```
Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.
```

**Constraints:**

```
1 <= nums.length <= 100
0 <= nums[i] <= 100
```

#### EXPLANATION:

easy的题目， 只要对当前的nums进行遍历， 如果当前的数字出现过偶数次， 那么就将formed数量+1， 最后将formed数量和 总数量 - formed * 2 返回就可以。

#### SOLUTION:

```
class Solution {
func numberOfPairs(_ nums: [Int]) -> [Int] {
var dic:Dictionary<Int,Int> = [:]
var formed:Int = 0
for num in nums {
if dic[num] == nil {
dic[num] = 1
} else {
dic[num] = dic[num]! + 1
}
if dic[num]! % 2 == 0 {
formed += 1
}
}
return [formed, nums.count - formed * 2]
}
}
```