[LeetCode] 59. Spiral Matrix II_Medium tag: array, DFS

Given a positive integer n, generate an n x n matrix filled with elements from 1 to n2 in spiral order.

 

Example 1:

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

Example 2:

Input: n = 1
Output: [[1]]

 

Constraints:

• 1 <= n <= 20

Ideas: 跟[LeetCode] 54. Spiral Matrix_Medium tag: array, DFS 的idea一样，只是将visited去掉，可以直接用返回的ans来标记是否visited过

T: O(m * n)      S: O(1)

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        self.n = n
        self.matrix = [[-1] * n for _ in range(n)]
        self.dirs = [(0, 1), (1, 0), ( 0, -1), (-1, 0)]
        self.helper(0, 0, 0, 1)
        return self.matrix
    
    
    
    def helper(self, i, j, dire, num):
        
        self.matrix[i][j] = num
        nr, nc = i + self.dirs[dire][0], j + self.dirs[dire][1]
        if 0 <= nr < self.n and 0 <= nc < self.n and self.matrix[nr][nc] == -1:
            self.helper(nr, nc, dire, num + 1)
        else:
            new_dir = (dire + 1) % 4
            nr, nc = i + self.dirs[new_dir][0], j + self.dirs[new_dir][1]
            if 0 <= nr < self.n and 0 <= nc < self.n and self.matrix[nr][nc] == -1:
                self.helper(nr, nc, new_dir, num + 1)

 

Code 2:

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        self.matrix = [[0]* n for _ in range(n)]
        self.n = n
        self.dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        self.dfs(0, 0, 0, 1)
        return self.matrix
    
    
    def dfs(self, i, j, dire, val):
        self.matrix[i][j] = val
        for k in range(4):
            new_dire = (dire + k)%4
            nr, nc = i + self.dirs[new_dire][0], j + self.dirs[new_dire][1]
            if 0 <= nr < self.n and 0 <= nc < self.n and self.matrix[nr][nc] == 0:
                self.dfs(nr, nc, new_dire, val + 1)