README.md

2326. Spiral Matrix IV

You are given two integers m and n, which represent the dimensions of a matrix.

You are also given the head of a linked list of integers.

Generate an m x n matrix that contains the integers in the linked list presented in spiral order (clockwise), starting from the top-left of the matrix. If there are remaining empty spaces, fill them with -1.

Return the generated matrix.

Example 1:

Input: m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
Output: [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
Explanation: The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.

Example 2:

Input: m = 1, n = 4, head = [0,1,2]
Output: [[0,1,2,-1]]
Explanation: The diagram above shows how the values are printed from left to right in the matrix.
The last space in the matrix is set to -1.

Constraints:

• 1 <= m, n <= 105
• 1 <= m * n <= 105
• The number of nodes in the list is in the range [1, m * n].
• 0 <= Node.val <= 1000

Solutions (Python)

1. Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]:
        curr = head
        i, j = 0, -1
        ret = [[-1] * n for _ in range(m)]

        while curr is not None:
            for (a, b) in [(0, 1)] * n + [(1, 0)] * (m - 1) + [(0, -1)] * (n - 1) + [(-1, 0)] * (m - 2):
                if curr is None:
                    break

                i += a
                j += b
                ret[i][j] = curr.val
                curr = curr.next

            m -= 2
            n -= 2

        return ret