[LeetCode] 1029. Two City Scheduling
作者:互联网
There are 2N
people a company is planning to interview. The cost of flying the i
-th person to city A
is costs[i][0]
, and the cost of flying the i
-th person to city B
is costs[i][1]
.
Return the minimum cost to fly every person to a city such that exactly N
people arrive in each city.
Example 1:
Input: [[10,20],[30,200],[400,50],[30,20]] Output: 110 Explanation: The first person goes to city A for a cost of 10. The second person goes to city A for a cost of 30. The third person goes to city B for a cost of 50. The fourth person goes to city B for a cost of 20. The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.
Note:
1 <= costs.length <= 100
- It is guaranteed that
costs.length
is even. 1 <= costs[i][0], costs[i][1] <= 1000
两地调度。提议是给你一个长度为2N的二维数组代表有2N个人需要去A,B两地面试的路费cost,请做一下安排使得AB两地面试人数相同且路费最小。
思路是贪心,我这里直接翻译一下LC discussion的最高票答案好了,同时参考题干中的例子。
costs = [[10,20],[30,200],[400,50],[30,20]]
首先假设,如果把所有的人都安排去A地参加面试,那么会得到cost = 10 + 30 + 400 + 30 = 470。但是因为题目要求需要送一半的人去B地面试,那么现在需要选择送什么人去B地呢?可以这样想,既然我目前已经花了470块送所有人去A地面试,那么我每送一个人去B地我都算是赚到了,因为我会得到来自A地的退款。按照这个思路,应该是去检查如何安排才能使退款更多。计算退款refund的方式是
refund[i] = cost[i][1] - cost[i][0]
得到数组refund = [10, 170, -350, -10]
这里每个数字,如果是正数则说明我们需要额外付钱,如果是负数说明可以得到退款。
接下来对refund数组排序,得到refund = [-350, -10, 10, 170],如果按照这个逻辑去安排,则最小花费是一开始的470 - 350 - 10 = 110块。
时间O(nlogn) - sort
空间O(n)
Java实现
1 class Solution { 2 public int twoCitySchedCost(int[][] costs) { 3 int N = costs.length / 2; 4 int[] refund = new int[N * 2]; 5 int minCost = 0; 6 int index = 0; 7 for (int[] cost : costs) { 8 refund[index++] = cost[1] - cost[0]; 9 minCost += cost[0]; 10 } 11 Arrays.sort(refund); 12 for (int i = 0; i < N; i++) { 13 minCost += refund[i]; 14 } 15 return minCost; 16 } 17 }
标签:refund,City,10,int,30,1029,cost,Scheduling,city 来源: https://www.cnblogs.com/cnoodle/p/13043535.html