[leetcode] 1029. Two City Scheduling

[leetcode] 1029. Two City Scheduling

Description

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

Example 1:

Input: costs = [[10,20],[30,200],[400,50],[30,20]]Output: 110Explanation: 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.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]Output: 3086

Constraints:

2 * n == costs.length2 <= costs.length <= 100costs.length is even.1 <= aCosti, bCosti <= 1000

分析

题目的意思是：有2n个人，数组的第i项表示的是第i个人去城市a的代价aCosti和去城市b的代价bCosti，用数组表示。

我参考了一下答案的思路，很巧妙：

按照 aCosti - bCosti 从小到大排序；将前 n个人飞往 A 市，其余人飞往 B 市，并计算出总费用。

思路是每个数组按照两项相减进行排序，然后取前n个人飞往A城市，其余的人飞往B城市。

代码

class Solution: def twoCitySchedCost(self, costs: List[List[int]]) -> int: res=0 costs.sort(key=lambda x:x[0]-x[1]) n=len(costs) for i in range(n): if(i

参考文献

​​两地调度​​

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