889. Spiral Matrix III

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

正常就是模拟，discuss有很强悍的解法

Intuition:
Take steps one by one.
If the location is inside of grid, add it to res.
But how to simulate the path?

It seems to be annoying, but if we oberserve the path:

move right 1 step, turn right
move down 1 step, turn right
move left 2 steps, turn right
move top 2 steps, turn right,
move right 3 steps, turn right
move down 3 steps, turn right
move left 4 steps, turn right
move top 4 steps, turn right,

we can find the sequence of steps: 1,1,2,2,3,3,4,4,5,5....

So there are two thing to figure out:

how to generate sequence 1,1,2,2,3,3,4,4,5,5
how to turn right?

Generate sequence 1,1,2,2,3,3,4,4,5,5
Let n be index of this sequence.
Then A0 = 1, A1 = 1, A2 = 2 ......
We can include that An = n / 2 + 1

How to turn right?
By cross product:
Assume current direction is (x, y) in plane, which is (x, y, 0) in space.
Then the direction after turn right (x, y, 0) × (0, 0, 1) = (y, -x, 0)
Translate to code: tmp = x; x = y; y = -tmp;

By arrays of arrays:
The directions order is (0,1),(1,0),(0,-1),(-1,0), then repeat.
Just define a variable.

Time Complexity:
O(max(M,N) ^ 2)

vector<vector<int>> spiralMatrixIII(int R, int C, int r, int c) {
        vector<vector<int>> res = {{r, c}};
        int x = 0, y = 1, tmp;
        for (int n = 0; res.size() < R * C; n++) {
            for (int i = 0; i < n / 2 + 1; i++) {
                r += x, c += y;
                if (0 <= r && r < R && 0 <= c && c < C)
                    res.push_back({r, c});
            }
            tmp = x, x = y, y = -tmp;
        }
        return res;
    }

class Solution:
    def spiralMatrixIII(self, R, C, r, c):
        """
        :type R: int
        :type C: int
        :type r0: int
        :type c0: int
        :rtype: List[List[int]]
        """
        res=[[r,c]]
        x,y,n=0,1,0
        while len(res)<R*C:
            for _ in range(n//2+1):
                r,c=r+x,c+y
                if 0<=r<R and 0<=c<C: res.append([r,c])
            x,y=y,-x
            n+=1
        return res
    
s=Solution()
print(s.spiralMatrixIII(R = 1, C = 4, r = 0, c = 0))

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