1029. Two City Scheduling

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

class Solution {
     public int twoCitySchedCost(int[][] costs) {
        Arrays.sort(costs, (a, b) -> {
            return (a[0] - a[1]) - (b[0] - b[1]);
        });
        
        int res = 0;
        for (int i = 0;i<costs.length;i++) {
            if (i < costs.length/2) {
                res += costs[i][0];
            } else 
                res += costs[i][1];
        }
        
        return res;
    }
}

一开始以为简单，结果想了半天也没做出来，倒是想到了给difference排序，但是为啥排序和怎么用没想明白。

后来想明白了：如果每一项都只选最小的那肯定会出问题，不满足每个city有N/2人。所以还需要考虑到两个的差值，先对差值排序，然后选N/2个第一个，剩下一半选第二个。

https://leetcode.com/problems/two-city-scheduling/discuss/278898/Java-2ms-sorting-solution-with-explanation

Greedy思想