leetcode讲解--885. 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

题目地址

讲解

这道题花了我很多时间，主要在于没有找到规律。螺线中有一条很重要的规律，臂长的增加是有规律的：1 1 2 2 3 3 ...

而后就是结束条件如何设定，这里我使用了堆满结果空间就返回的方式。

Java代码

class Solution {
    public int[][] spiralMatrixIII(int R, int C, int r0, int c0) {
        int[][] result = new int[R*C][2];
        int count=0;
        result[0] = new int[]{r0, c0};
        int size=0;
        while(count<R*C-1){
            size++;
            for(int j=1;j<=size;j++){
                c0 = c0+1;
                if(r0>=0 && c0>=0 && r0<R && c0<C){
                    result[++count] = new int[]{r0, c0};
                    if(count==R*C-1){
                        return result;
                    }
                }
            }
            for(int j=1;j<=size;j++){
                r0=r0+1;
                if(r0>=0 && c0>=0 && r0<R && c0<C){
                    result[++count] = new int[]{r0, c0};
                    if(count==R*C-1){
                        return result;
                    }
                }
            }
            size++;
            for(int j=1;j<=size;j++){
                c0 = c0-1;
                if(r0>=0 && c0>=0 && r0<R && c0<C){
                    result[++count] = new int[]{r0, c0};
                    if(count==R*C-1){
                        return result;
                    }
                }
            }
            for(int j=1;j<=size;j++){
                r0=r0-1;
                if(r0>=0 && c0>=0 && r0<R && c0<C){
                    result[++count] = new int[]{r0, c0};
                    if(count==R*C-1){
                        return result;
                    }
                }
            }
        }
        return result;
    }
}