Spiral Matrix IV - Solution & Explanation

This approach traverses the matrix in a spiral order by systematically altering the direction when encountering the boundary or an already filled space. Start with the initial direction as 'right'. Change direction to 'down', 'left', 'up' as necessary when a boundary or an already filled cell is encountered.

The provided Python code defines a function spiralMatrix that takes the dimensions m, n and the head of a linked list. It begins by creating a m x n matrix initialized with -1. The spiral movement is driven by the dirs array, which holds possible movement directions. For each direction change, we ensure it keeps within valid boundaries or filled cell conditions. This implementation guarantees that the matrix is filled in a proper spiral order, transitioning directions when necessary.

In this approach, we fill the matrix in a spiral order by completing one 'layer' of the spiral at a time. Start filling from the outer layer to the inner layers progressively until the entire matrix is filled. Each layer comprises four segments: top row, right column, bottom row, and left column.

The Java implementation manages a layer by tracking the limits of the current upward, downward, leftward, and rightward possible moves. The solution fills all available cells in these 'layers', transitioning from one direction to the next only after fully exhausting entries along the current path, and proceeds inward. The method completes filling the complete list coupled with the matrix traversal ensuring optimal manipulation through nested loops guarding respective limits.

We define a two-dimensional array ans to store the elements in the linked list, initially all filled with -1. We define three variables i, j, k, representing the current row, column, and direction respectively. We define an array dirs to represent the offsets of the four directions.

Then we start traversing the linked list. Each time we traverse a node, we fill the current node's value into ans[i][j], then update the linked list pointer. If the linked list is empty, it means all elements have been filled and we exit the loop.

Otherwise, we need to find the position of the next element. We can calculate the next position (x, y) through the current position (i, j) and the current direction k. If (x, y) is within the range of the matrix, and ans[x][y] is -1, it means (x, y) has not been filled yet, so we take (x, y) as the next position. Otherwise, we need to change the direction.

After traversing the linked list, we get a spiral matrix and return it.

The time complexity is O(m times n), and the space complexity is O(m times n), where m and n represent the number of rows and columns of the matrix, respectively.