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885. 螺旋矩阵 III

题目描述

在 rows x cols 的网格上，你从单元格 (rStart, cStart) 面朝东面开始。网格的西北角位于第一行第一列，网格的东南角位于最后一行最后一列。

你需要以顺时针按螺旋状行走，访问此网格中的每个位置。每当移动到网格的边界之外时，需要继续在网格之外行走（但稍后可能会返回到网格边界）。

最终，我们到过网格的所有 rows x cols 个空间。

按照访问顺序返回表示网格位置的坐标列表。

 

示例 1：

输入：rows = 1, cols = 4, rStart = 0, cStart = 0
输出：[[0,0],[0,1],[0,2],[0,3]]

示例 2：

输入：rows = 5, cols = 6, rStart = 1, cStart = 4
输出：[[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]]

 

提示：

  • 1 <= rows, cols <= 100
  • 0 <= rStart < rows
  • 0 <= cStart < cols

解法

方法一

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class Solution:
    def spiralMatrixIII(
        self, rows: int, cols: int, rStart: int, cStart: int
    ) -> List[List[int]]:
        ans = [[rStart, cStart]]
        if rows * cols == 1:
            return ans
        k = 1
        while True:
            for dr, dc, dk in [[0, 1, k], [1, 0, k], [0, -1, k + 1], [-1, 0, k + 1]]:
                for _ in range(dk):
                    rStart += dr
                    cStart += dc
                    if 0 <= rStart < rows and 0 <= cStart < cols:
                        ans.append([rStart, cStart])
                        if len(ans) == rows * cols:
                            return ans
            k += 2

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class Solution {
    public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
        int cnt = rows * cols;
        int[][] ans = new int[cnt][2];
        ans[0] = new int[] {rStart, cStart};
        if (cnt == 1) {
            return ans;
        }
        for (int k = 1, idx = 1;; k += 2) {
            int[][] dirs = new int[][] {{0, 1, k}, {1, 0, k}, {0, -1, k + 1}, {-1, 0, k + 1}};
            for (int[] dir : dirs) {
                int r = dir[0], c = dir[1], dk = dir[2];
                while (dk-- > 0) {
                    rStart += r;
                    cStart += c;
                    if (rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols) {
                        ans[idx++] = new int[] {rStart, cStart};
                        if (idx == cnt) {
                            return ans;
                        }
                    }
                }
            }
        }
    }
}

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class Solution {
public:
    vector<vector<int>> spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
        int cnt = rows * cols;
        vector<vector<int>> ans;
        ans.push_back({rStart, cStart});
        if (cnt == 1) return ans;
        for (int k = 1;; k += 2) {
            vector<vector<int>> dirs = {{0, 1, k}, {1, 0, k}, {0, -1, k + 1}, {-1, 0, k + 1}};
            for (auto& dir : dirs) {
                int r = dir[0], c = dir[1], dk = dir[2];
                while (dk-- > 0) {
                    rStart += r;
                    cStart += c;
                    if (rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols) {
                        ans.push_back({rStart, cStart});
                        if (ans.size() == cnt) return ans;
                    }
                }
            }
        }
    }
};

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func spiralMatrixIII(rows int, cols int, rStart int, cStart int) [][]int {
    cnt := rows * cols
    ans := [][]int{[]int{rStart, cStart}}
    if cnt == 1 {
        return ans
    }
    for k := 1; ; k += 2 {
        dirs := [][]int{{0, 1, k}, {1, 0, k}, {0, -1, k + 1}, {-1, 0, k + 1}}
        for _, dir := range dirs {
            r, c, dk := dir[0], dir[1], dir[2]
            for dk > 0 {
                rStart += r
                cStart += c
                if rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols {
                    ans = append(ans, []int{rStart, cStart})
                    if len(ans) == cnt {
                        return ans
                    }
                }
                dk--
            }
        }
    }
}

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