标签:dirs Medium Matrix helper 54 self ans visited matrix
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)
标签:dirs,Medium,Matrix,helper,54,self,ans,visited,matrix 来源: https://www.cnblogs.com/Johnsonxiong/p/15171212.html
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