LeetCode: Spiral Matrix Solution

Approach

Recursively go with the order of Right - Down - Left - Up

Implementation

1var spiralOrder = function (matrix) {
2  const [m, n] = [matrix.length, matrix[0].length]
3  const visited = Array.from({ length: m }, _ => Array(n).fill(false))
4  const res = []
5  const DIR = {
6    R: 0,
7    D: 1,
8    L: 2,
9    U: 3,
10  }
11
12  const isValidCell = (i, j) =>
13    0 <= i && i < m && 0 <= j && j < n && !visited[i][j]
14
15  const recursion = (i, j, dir) => {
16    if (!isValidCell(i, j)) {
17      dir === DIR.R &&
18        isValidCell(i + 1, j - 1) &&
19        recursion(i + 1, j - 1, DIR.D)
20      dir === DIR.D &&
21        isValidCell(i - 1, j - 1) &&
22        recursion(i - 1, j - 1, DIR.L)
23      dir === DIR.L &&
24        isValidCell(i - 1, j + 1) &&
25        recursion(i - 1, j + 1, DIR.U)
26      dir === DIR.U &&
27        isValidCell(i + 1, j + 1) &&
28        recursion(i + 1, j + 1, DIR.R)
29      return
30    }
31
32    res.push(matrix[i][j])
33    visited[i][j] = true
34
35    dir === DIR.R && recursion(i, j + 1, dir)
36    dir === DIR.D && recursion(i + 1, j, dir)
37    dir === DIR.L && recursion(i, j - 1, dir)
38    dir === DIR.U && recursion(i - 1, j, dir)
39  }
40
41  recursion(0, 0, DIR.R)
42
43  return res
44}

References

Original problem

Similar problems

Spiral Matrix II

Spiral Matrix III

Spiral Matrix IV