leetcode 1029. Two City Scheduling 解法 python

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最新推荐文章于 2025-06-04 15:38:55 发布

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本文介绍了一个面试调度问题的解决方案，目标是最小化2N个人分别前往两个城市的总成本，确保每个城市都有N人到达。通过计算每个人前往A城市与B城市的成本差值，并对这些差值进行排序，可以有效地找到最小成本的方案。

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一.问题描述

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

二.解决思路

用第一个维度减第二个维度

得到去a城市和去b城市两者的差值

差值负得越多，说明去a城市比去b城市越划算，正得越多则反之

将差值排序，然后每次加一个正得对多的和负得最多得，最后就能得到mincost

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三.源码

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        N = len(costs)
        diff = [c[0] - c[1] for c in costs]
        indices =  sorted(range(0,N), key=lambda k:diff[k])
        result = 0
        for i in range(int(N/2)):
            result += costs[indices[i]][0]
            result += costs[indices[N-i-1]][1]
        return result