1030. Matrix Cells in Distance Order

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

 

Example 1:

Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]
Example 2:

Input: R = 2, C = 2, r0 = 0, c0 = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.
Example 3:

Input: R = 2, C = 3, r0 = 1, c0 = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].
 

Note:

1 <= R <= 100
1 <= C <= 100
0 <= r0 < R
0 <= c0 < C

这题主要让我们把某个给定点周围所有的点进行一个排序，按照与它的距离来排序。一开始想着一圈一圈遍历，后来发现太麻烦了，直接硬破，全部加进去然后进行一次排序。

class MatrixCellsInDistanceOrder : public Solution {
public:
    void Exec() {

    }
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
        vector<vector<int>> res;
        
        for (int r = 0; r < R; r++) {
            for (int c = 0; c < C; c++) {
                res.push_back(vector<int>{r, c});
            }
        }
        std::sort(res.begin(), res.end(), [r0, c0](vector<int> &l, vector<int> &r) -> bool {
            return (abs(l[0] - r0) + abs(l[1] - c0)) < (abs(r[0] - r0) + abs(r[1] - c0));
        });

        return res;
    }
};