LeetCode 885. Spiral Matrix III

原题链接在这里：https://leetcode.com/problems/spiral-matrix-iii/

题目：

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

题解：

At the beginning, turn at 1 and 1, then 2 and 2, then 3 and 3, and goes on.

Thus, we need to figure out the loop relationship.

Every 2 times, we goes into the repeated steps. When finishing repeated steps, turn.

After 2 times, increase repeated steps.

With this, we know 2 times loop is outside repeated steps loop.

Time Complexity: O(R * C).

Space: O(1). regardless res.

AC Java:

class Solution {
    int [][] dirs = new int[][]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    
    public int[][] spiralMatrixIII(int R, int C, int r0, int c0) {
        int [][] res = new int[R * C][2];
        if(R == 0 || C == 0){
            return res;
        }
        
        res[0] = new int[]{r0, c0};
        
        int count = 1;
        int index = 0;
        int k = 1;
        while(count < R * C){
            for(int j = 0; j<2; j++){
                int [] dir = dirs[index % 4];
                for(int i = 0; i<k; i++){
                    r0 += dir[0];
                    c0 += dir[1];
                    if(r0 >= 0 && r0 < R && c0 >=0 && c0 < C){
                        res[count++] = new int[]{r0, c0};
                    }
                }
                
                index++;
            }
            
            k++;
        }
        
        return res;
     }
}

类似Spiral Matrix, Spiral Matrix II.