Problem
On a 2 dimensional grid with R rows and C columns, we start at (r0, c0) facing east.
Here, the north-west corner of the grid is at the first row and column, and the south-east corner of the grid is at the last row and column.
Now, we walk in a clockwise spiral shape to visit every position in this grid.
Whenever we would move outside the boundary of the grid, we continue our walk outside the grid (but may return to the grid boundary later.)
Eventually, we reach all R * C spaces of the grid.
Return a list of coordinates representing the positions of the grid in the order they were visited.
Example 1:
Input: R = 1, C = 4, r0 = 0, c0 = 0 Output: [[0,0],[0,1],[0,2],[0,3]]
Example 2:
Input: R = 5, C = 6, r0 = 1, c0 = 4 Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]
Note:
1 <= R <= 1001 <= C <= 1000 <= r0 < R0 <= c0 < C
Solution: Simulation
We can find out the moving sequence is ESWWNNEEESSSWWWWNNNN.
The pattern is 1,1,2,2,3,3,4,4,… steps in one direction, and turn right for 90 degrees.
directions are E,S,W,N,E,S,W,N…
Time complexity: O(max(R,C)^2)
Space complexity: O(1) or O(RC) if ans included.
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// Author: Huahua // Running time: 52 ms class Solution { public: vector<vector<int>> spiralMatrixIII(int R, int C, int r0, int c0) { vector<vector<int>> ans; int k = 0; int dirs[][2]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; // ESWN int d = 0; // E ans.push_back({r0, c0}); if (ans.size() == R * C) return ans; while (++k) { for (int i = 0; i < 2; ++i) { for (int j = 0; j < k; ++j) { c0 += dirs[d][0]; r0 += dirs[d][1]; if (c0 < 0 || c0 >= C || r0 < 0 || r0 >= R) continue; ans.push_back({r0, c0}); if (ans.size() == R * C) return ans; } d = (d + 1) % 4; } } return {}; } }; |