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]].
二维矩阵给定其中一个点，问它到其它所有点距离从小到大排序。
Thoughts
从小到大排序也就是从那个点开始一层层往外扩，也就是BFS.
class Solution {
public:
vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
vector<pair<int, int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
vector<vector<int>> res;
queue<pair<int, int>> q;
q.push({r0, c0});
res.push_back({r0, c0});
vector<vector<int>> v(R, vector<int>(C));
v[r0][c0] = true;
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
auto t = q.front(); q.pop();
for (const auto dir : dirs) {
int x = t.first + dir.first;
int y = t.second + dir.second;
string k = to_string(x) + ":" + to_string(y);
if (x >= 0 && x < R && y >= 0 && y < C && !v[x][y]) {
res.push_back({x, y});
v[x][y] = true;
q.push({x, y});
}
}
}
}
return res;
}
};