885. Spiral Matrix III

Posted 2021-01-14 bernieloveslife

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

class Solution:
    def spiralMatrixIII(self, R, C, r0, c0):
        """
        :type R: int
        :type C: int
        :type r0: int
        :type c0: int
        :rtype: List[List[int]]
        """
        count = 1
        l = 1
        res = []
        direction = 1
        res.append([r0,c0])
        while count<R*C:
            if direction==1:
                c0 += l
                # print(‘direction:1,pos:‘,r0,c0)
                for i in range(c0-l+1,c0+1): #不考虑起点，考虑终点
                    if 0<=r0<R and 0<=i<C:
                        res.append([r0,i])
                        count += 1
                #         print(count)
                # print(res)
                direction = 2
                continue
            if direction==2:
                r0 += l
                # print(‘direction:2,pos:‘,r0,c0)
                for i in range(r0-l+1,r0+1):
                    if 0<=i<R and 0<=c0<C:
                        res.append([i,c0])
                        count += 1
                #         print(count)
                # print(res)
                direction = 3
                l += 1
                continue
            if direction==3:
                c0 -= l
                # print(‘direction:3,pos:‘,r0,c0)
                for i in range(c0+l-1,c0-1,-1):
                    if 0<=r0<R and 0<=i<C:
                        res.append([r0,i])
                        count += 1
                #         print(count)
                # print(res)
                direction = 4
                continue
            if direction == 4:
                r0 -= l
                # print(‘direction:4,pos:‘,r0,c0)
                for i in range(r0+l-1,r0-1,-1):
                    if 0<=i<R and 0<=c0<C:
                        res.append([i,c0])
                        count += 1
                #         print(count)
                # print(res)
                direction = 1
                l += 1
                continue
        return res

每次转换方向，每走两次长度加1，循环跳出条件为数量达到所有格子数。对于每一次走，判断若在棋盘范围内就保存下来。