LeetCode 1029. Two City Scheduling

Posted on | In |
Words count in article: 231 | Reading time ≈ 1

题目

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:

1
2
3
4
5
6
7
8
9
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

思路

Greedy.

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
class Solution(object):
    def twoCitySchedCost(self, costs):
        """
        :type costs: List[List[int]]
        :rtype: int
        """
        diff = []
        for i, cost in enumerate(costs):
            diff.append((cost[0]-cost[1], i))
        diff.sort()
        res, cnt = 0, 0
        for i, (j, k) in enumerate(diff):
            if i < len(costs)/2:
                res += costs[k][0]
            else:
                res += costs[k][1]
        return res