Given an m x n matrix, return all elements of the matrix in spiral order.

 

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]

Example 2:

Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]

 

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 10
• -100 <= matrix[i][j] <= 100

Ideas:  T: O(m * n),  S: O( m * n)

1. 建一个helper function， 方向为（right，down，left， up）， 先从right开始，不停往走，直到走到头或者已经visited过了，再换方向

2. 返回ans

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        self.matrix = matrix
        self.m = len(matrix)
        self.n = len(matrix[0])
        ans = []
        visited = set()
        self.dirs = [(0, 1), (1, 0), ( 0, -1), (-1, 0)]
        self.helper(0, 0, 0, ans, visited)
        return ans
    
    def helper(self, i, j, dire, ans, visited):
        
        visited.add((i, j))
        ans.append(self.matrix[i][j])
        nr, nc = i + self.dirs[dire][0], j + self.dirs[dire][1]
        if 0 <= nr < self.m and 0 <= nc < self.n and (nr, nc) not in visited:
            self.helper(nr, nc, dire, ans, visited)
        else:
            new_dir = (dire + 1) % 4
            nr, nc = i + self.dirs[new_dir][0], j + self.dirs[new_dir][1]
            if 0 <= nr < self.m and 0 <= nc < self.n and (nr, nc) not in visited:
                self.helper(nr, nc, new_dir, ans, visited)