[LeetCode] 1030. Matrix Cells in Distance Order 距離順序排列矩陣單元格

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. 1 <= R <= 100
2. 1 <= C <= 100
3. 0 <= r0 < R
4. 0 <= c0 < C

這道題給了一個R行C列的矩陣，又給了一個起始點 (r0, c0)，讓按照離起始點的曼哈頓距離從小到大排序坐標點。博主最先想到的方法就是從起始點開始進行廣度優先遍歷 Breadth-First Search，這樣保證了離起始點的距離是從小到大的，在遍歷的過程中將坐標加入結果 res 即可，寫法就是最普通的 BFS 遍歷，沒有太多需要注意的地方，參見代碼如下：

解法一：

class Solution {
public:
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
        vector<vector<int>> res;
        set<vector<int>> visited;
        vector<vector<int>> dirs{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        queue<vector<int>> q;
        q.push({r0, c0});
        visited.insert({r0, c0});
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            res.push_back(t);
            for (auto dir : dirs) {
                int x = t[0] + dir[0], y = t[1] + dir[1];
                if (x < 0 || x >= R || y < 0 || y >= C || visited.count({x, y})) continue;
                visited.insert({x, y});
                q.push({x, y});
            }
        }
        return res;
    }
};

其實我們並不用寫個 BFS 那么麻煩，直接自定義一個 comparator 給 res 數組重新排序即可，自定義的 comparator 要把 (r0, c0) 當參數傳進去，因為要求和其的曼哈頓距離，參見代碼如下：

解法二：

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](vector<int>& a, vector<int>& b) {
            return abs(a[0] - r0) + abs(a[1] - c0) < abs(b[0] - r0) + abs(b[1] - c0);
        });
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1030

參考資料：

https://leetcode.com/problems/matrix-cells-in-distance-order/

https://leetcode.com/problems/matrix-cells-in-distance-order/discuss/278843/O(N)-Java-BFS

https://leetcode.com/problems/matrix-cells-in-distance-order/discuss/278807/c%2B%2B-sorting-min-heap-solutions

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