Two City Scheduling Solutions in C++

// Source : https://leetcode.com/problems/two-city-scheduling/
// Author : Hao Chen
// Date   : 2019-04-21




class Solution {
private:
    static int diff(vector<int>& x) {
        return x[1] - x[0];
    }
    static bool cmpfunc(vector<int>& lhs, vector<int>& rhs) {
        return diff(lhs) > diff(rhs);
    }

public:
    // Just simply sort the array by comparing the different cost go to A city and B city
    // then the bigger difference would be in left and right side, and the smaller difference would be in the middle
    // We could simply let the first half go to A city, and the second half go to B city.
    int twoCitySchedCost(vector<vector<int>>& costs) {
        sort(costs.begin(), costs.end(), cmpfunc);
        int result = 0;
        int len = costs.size();
        for (int i=0; i<len/2; i++) {
            result += (costs[i][0] + costs[len-i-1][1]);
        }
        return result;
    }
};

/// Source : https://leetcode.com/problems/two-city-scheduling/
/// Author : liuyubobobo
/// Time   : 2019-04-20

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;


/// Sorting
/// Time Complexity: O(nlogn)
/// Space Complexity: O(1)
class Solution {
public:
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {

        vector<vector<int>> res;
        for(int i = 0; i < R; i ++)
            for(int j = 0; j < C; j ++)
               res.push_back({i, j});

        sort(res.begin(), res.end(), [r0, c0](const vector<int>& v1, const vector<int>& v2){

            int dis1 = abs(v1[0] - r0) + abs(v1[1] - c0);
            int dis2 = abs(v2[0] - r0) + abs(v2[1] - c0);
            return dis1 < dis2;
        });
        return res;
    }
};


int main() {

    return 0;
}