【LeetCode】 1029. Two City Scheduling

Description

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.

Note:

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

一家公司預計有2N個人要來面試。第i個人要飛到城市A的費用為costs[i][0]，而第i個人要飛到城市b的費用為costs[i][1]。

回傳每個城市都剛好有N個人時，飛行所需要的最小花費。

注意：

• 1 <= costs.length <= 100
• 保證 costs.length 是偶數。
• 1 <= costs[i][0], costs[i][1] <= 1000

Example:

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.

Solution

• 這題的重點在於要如何分配哪N個人要到城市A、哪N個人要到城市B。
• 我們將costs重新排列，依據[0] - [1]。
  • 以範例來說，會得到-10, -170, 350, 10，然後重新排序。
• 這個排序過後的陣列，越前面的就代表選A越省；反之越後面選B越省。
• 那麼我們就從中間切一半，前半選A、後半選B，計算總合即可。

Code

bool myCompare(vector<int> a, vector<int> b)
{
    return a[0] - a[1] < b[0] - b[1];
}

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int sum = 0;
        sort(costs.begin(), costs.end(), myCompare);
        for(int i = 0; i < costs.size() / 2; i++)
        {
            sum += costs[i][0] + costs[costs.size() - 1 - i][1];
        }
        return sum;
    }
};

tags: LeetCode C++