【每日一题Day73】LC2037使每位学生都有座位的最少移动次数

该博客讨论了如何在给定座位和学生位置的情况下，通过最小化移动次数使每个学生都能坐在座位上。解决方案是首先对座位和学生的位置进行排序，然后计算每个学生到达最近座位所需的移动次数，并累加得到总移动次数。这是一个关于算法和数据结构的问题，主要涉及排序和数学优化。

使每位学生都有座位的最少移动次数【LC2037】

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.

元旦快乐啦

思路：要使总移动次数最少，那么要将每个学生移动至离其最近的座位，因此将座位和学生的位置进行升序排序，每个学生需要的移动次数即为对应位置相减，累加返回最终结果

实现

class Solution {
    public int minMovesToSeat(int[] seats, int[] students) {
        Arrays.sort(seats);
        Arrays.sort(students);
        int res = 0;
        for (int i = 0; i < seats.length; i++){
            res += Math.abs(seats[i] - students[i]);
        }
        return res;

    }
}

复杂度
- 时间复杂度： $O (n)$
- 空间复杂度： $O (1)$

实现

class Solution {
    public int minMovesToSeat(int[] seats, int[] students) {
        Arrays.sort(seats);
        Arrays.sort(students);
        int res = 0;
        for (int i = 0; i < seats.length; i++){
            res += Math.abs(seats[i] - students[i]);
        }
        return res;

    }
}