#### QUESTION:

There are n seats and n students in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student.

You may perform the following move any number of times:

Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x - 1) Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat.

Note that there may be multiple seats or students in the same position at the beginning.

**Example 1:**

```
Input: seats = [3,1,5], students = [2,7,4]
Output: 4
Explanation: The students are moved as follows:
- The first student is moved from from position 2 to position 1 using 1 move.
- The second student is moved from from position 7 to position 5 using 2 moves.
- The third student is moved from from position 4 to position 3 using 1 move.
In total, 1 + 2 + 1 = 4 moves were used.
```

**Example 2:**

```
Input: seats = [4,1,5,9], students = [1,3,2,6]
Output: 7
Explanation: The students are moved as follows:
- The first student is not moved.
- The second student is moved from from position 3 to position 4 using 1 move.
- The third student is moved from from position 2 to position 5 using 3 moves.
- The fourth student is moved from from position 6 to position 9 using 3 moves.
In total, 0 + 1 + 3 + 3 = 7 moves were used.
```

**Example 3:**

```
Input: seats = [2,2,6,6], students = [1,3,2,6]
Output: 4
Explanation: Note that there are two seats at position 2 and two seats at position 6.
The students are moved as follows:
- The first student is moved from from position 1 to position 2 using 1 move.
- The second student is moved from from position 3 to position 6 using 3 moves.
- The third student is not moved.
- The fourth student is not moved.
In total, 1 + 3 + 0 + 0 = 4 moves were used.
```

**Constraints:**

```
n == seats.length == students.length
1 <= n <= 100
1 <= seats[i], students[j] <= 100
```

#### EXPLANATION:

这个就比较简单了. 用贪心的思想去做一下就可以. 将两个数组sort一下. 那么每个位置对应的则是最小的步数. 加起来就是结果了.

#### SOLUTION:

```
class Solution {
func minMovesToSeat(_ seats: [Int], _ students: [Int]) -> Int {
var seatsSorted:[Int] = seats.sorted()
var studentsSorted:[Int] = students.sorted()
var result:Int = 0
for i in 0...seats.count-1 {
result += abs(seatsSorted[i] - studentsSorted[i])
}
return result
}
}
```