June Leetcoding Challenge(6.3)

Two City Scheduling

Solution

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

思路：

先将所有的人送到A城市，计算出来minCost，计算到A城市和B城市之间的差值，在将这些差值放入一个PriorityQueue，将前N/2个最小的数与minCost相加得到最后的结果

代码：

class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int minCost = 0;
        int n = costs.length;
        for(int i=0;i<n;i++){
            minCost += costs[i][0];
        }
        
        PriorityQueue<Integer> minQ = new PriorityQueue<Integer>();
        for(int i=0;i<n;i++){
            minQ.add(costs[i][1]-costs[i][0]);
        }
        for(int i = 0;i<n/2;i++)
            minCost = minCost + minQ.poll();
        return minCost;
    }
}