leetcode1029. Two City Scheduling

Posted 2021-10-23 seyjs

题目如下：

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. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

解题思路：我的方法是把costs按照costs[i][0] - costs[i][1]的差值从小到大排序，然后前面一半的人去A，后面一半的人去B。至于为什么这样做，我也说不上来，感觉。

代码如下：

class Solution(object):
    def twoCitySchedCost(self, costs):
        """
        :type costs: List[List[int]]
        :rtype: int
        """        
        def cmpf(item1,item2):
            d1 = item1[0] - item1[1]
            d2 = item2[0] - item2[1]
            return d1 - d2

        costs.sort(cmp=cmpf)
        res = 0
        for i in range(len(costs)):
            if i < len(costs)/2:
                res += costs[i][0]
            else:
                res += costs[i][1]
        #print costs
        return res