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描述
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: 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.
Note:
n == seats.length == students.length
1 <= n <= 100
1 <= seats[i], students[j] <= 100
解析
根据题意，给出了有 n 个座位的列表 seats ，其中 seats[i] 表示的是座位号，同时又给出了有 n 个学生的列表 students ，其中 students[i] 表示的是该学生现在做的座位号，我们可以执行以下的操作无数次：
- 将一个学生移动到其座位号增加一个的位置或者减少一个的位置，比如将座位号为 3 的学生可以移动到座位号为 2 或者 4 的位置上
题目要求我们执行多少次上述操作，可以将 students 中的学生都安排到 seats 座位上，且没有两个学生坐在同一个座位上。其实思路很简单，就是个找规律题。
假如 seats = [12,14,19,19,12] ，students = [19,2,17,20,7] ，我们先经过排序可以得到：
seats = [12,12,14,19,19]
students = [2,7,17,19,20]
- 我们可以发现 students[0] 学生离得最近的座位号就是 seats[0] ，seats[1] 及其后面的座位肯定相等或者更远不做考虑，此时 seats[0] 就被占了
- students[1] 学生离得最近的座位号就是 seats[0] 和 seats[1]，
- seats[2] 及其后面的座位肯定相等或者更远不做考虑，但是 seats[0] 已经被占，所以离得最近的座位就是 seats[1]
- students[2] 学生离得最近的座位号就是 seats[1] 和 seats[2]，
- seats[3] 及其后面的座位肯定相等或者更远不做考虑，但是 seats[1] 已经被占，所以离得最近的座位就是 seats[2]
- 同样的规律继续进行下去我们发现，其实相同索引 i 的 seats[i] 的座位就是 students[i] 离的最近的位置，直接将 abs(seats[i]-students[i]) 加入结果中即可得到结果。
解答
class Solution(object):
def minMovesToSeat(self, seats, students):
"""
:type seats: List[int]
:type students: List[int]
:rtype: int
"""
result = 0
seats.sort()
students.sort()
save = []
for i in range(len(seats)):
result += abs(seats[i]-students[i])
return result
运行结果
Runtime: 61 ms, faster than 100.00% of Python online submissions for Minimum Number of Moves to Seat Everyone.
Memory Usage: 13.4 MB, less than 100.00% of Python online submissions for Minimum Number of Moves to Seat Everyone.
原题链接：leetcode.com/problems/mi…
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