LeetCode-Spiral Matrix III

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Description：
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

题意：给定一个二位数组的行数R和列数C，从点r0, c0开始按照时钟的走法遍历完整个二维数组，输出遍历的数组的路径；

解法：观察示例我们可以得出如下的规律

• 初始的方向为向东
• 包含东南西北四个前进方向
• 前进的步长从1开始递增
• 每前进两次当前步长时，步长增1

根据以上的规则，我们可以定义四个遍历的方向及当前方向下标位置，当前步长及步长增加的周期；在步长的增加周期中，我们每次朝当前方向前进步长的距离，并更新前进方向，结束步长周期步长增1，继续上述操作，直到遍历完数组中的所有元素；

Java

class Solution {
    public int[][] spiralMatrixIII(int R, int C, int r0, int c0) {
        int sumStep = R * C;
        int[][] result = new int[sumStep][2];
        int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int indexDir = 0;
        int times = 1;
        result[0][0] = r0;
        result[0][1] = c0;
        for (int i = 1; i < sumStep;) {
            int cycle = 2;
            while (cycle -- > 0) {
                for (int j = 0; j < times; j++) {
                    r0 += dir[indexDir][0];
                    c0 += dir[indexDir][1];
                    if (isValid(R, C, r0, c0)) {
                        result[i][0] = r0;
                        result[i][1] = c0;
                        i++;
                    }
                }
                indexDir = (indexDir + 1) % dir.length;
            }
            times++;
        }
        return result;
    }
    private boolean isValid(int R, int C, int r, int c) {
        if (r < 0 || r >= R || c < 0 || c >= C) return false;
        return true;
    }
}