leetcode讲解--885. Spiral Matrix III

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原文链接：https://segmentfault.com/a/1190000017527697

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#数据结构与算法 #java

本文深入探讨了一种螺旋矩阵遍历算法，该算法用于在二维网格上按顺时针螺旋形路径访问所有位置。文章详细解释了算法的实现原理，包括如何处理边界条件和螺旋臂长增长的规律，通过具体实例展示了算法的应用过程。

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

题目地址

讲解

这道题花了我不少时间，主要在于没有找到规律。螺线中有一条很重要的规律，臂长的增长是有规律的：1 1 2 2 3 3 ...

然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;
    }
}