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To traverse a rectangular matrix in clockwise spiral order, start at the upper-left, move across the top row, down the right edge, back across the bottom, and up the left edge. Repeat on the smaller rectangle inside. The key to avoiding duplicates is to stop when the remaining boundaries cross—especially when only one row or one column is left.
What spiral order means
LeetCode 54 defines the task as: “Given an m x n matrix, return all elements of the matrix in spiral order.” The traversal begins in the upper-left corner and follows the outside perimeter clockwise before moving inward. Its examples and constraints are in the LeetCode 54 problem statement.
For example, the matrix [[1,2,3],[4,5,6],[7,8,9]] produces [1,2,3,6,9,8,7,4,5]. Each cell appears once, so for a nonempty rectangular matrix the result contains m*n values.
Trace the traversal with a step-by-step visualizer
A useful scrubber should show the matrix, highlight the current cell and direction, and display the four boundaries that define the unvisited rectangle. The following trace uses the 3-by-4 example from LeetCode. At each step, the cell is appended to the result; after finishing an edge, its boundary moves inward.
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[[1,2,3,4],[5,6,7,8],[9,10,11,12]]
| Step | Current cell | Direction | Boundaries after step (top, bottom, left, right) | Result so far |
|---|---|---|---|---|
| 1 | (0,0) = 1 | Right | (0,2,0,3) | [1] |
| 2 | (0,1) = 2 | Right | (0,2,0,3) | [1,2] |
| 3 | (0,2) = 3 | Right | (0,2,0,3) | [1,2,3] |
| 4 | (0,3) = 4 | Right; top edge complete | (1,2,0,3) | [1,2,3,4] |
| 5 | (1,3) = 8 | Down | (1,2,0,3) | [1,2,3,4,8] |
| 6 | (2,3) = 12 | Down; right edge complete | (1,2,0,2) | [1,2,3,4,8,12] |
| 7 | (2,2) = 11 | Left | (1,2,0,2) | [1,2,3,4,8,12,11] |
| 8 | (2,1) = 10 | Left | (1,2,0,2) | [1,2,3,4,8,12,11,10] |
| 9 | (2,0) = 9 | Left; bottom edge complete | (1,1,0,2) | [1,2,3,4,8,12,11,10,9] |
| 10 | (1,0) = 5 | Up; left edge complete | (1,1,1,2) | [1,2,3,4,8,12,11,10,9,5] |
| 11 | (1,1) = 6 | Right | (1,1,1,2) | [1,2,3,4,8,12,11,10,9,5,6] |
| 12 | (1,2) = 7 | Right; inner row complete | (2,1,1,2) | [1,2,3,4,8,12,11,10,9,5,6,7] |
The final boundaries have crossed: top is greater than bottom. There are no unvisited cells, and the result is [1,2,3,4,8,12,11,10,9,5,6,7]. A working interactive visualizer should update the highlighted cell, arrow, boundaries, and result together on every scrubber step; this trace describes the algorithm, not a tested online tool.
Use shrinking boundaries for a direct implementation
Maintain the edges of the unvisited rectangle: top, bottom, left, and right. Traverse its top, right, bottom, and left edges in that order, moving the matching boundary inward after each edge.
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- Set
top = 0,bottom = m - 1,left = 0, andright = n - 1. - While
top <= bottomandleft <= right, append the top row fromleftthroughright, then incrementtop. - If
top <= bottom, append the right column from the newtopthroughbottom, then decrementright. - If
left <= right, append the bottom row fromrightdown throughleft, then decrementbottom. - If
top <= bottom, append the left column frombottomup throughtop, then incrementleft.
The checks before the bottom-row and left-column traversals matter: earlier edges may already have consumed the last remaining row or column. Without these guards, the algorithm can append cells twice. This boundary-by-boundary method and the single-row and single-column caution are also described in the LeetCode community discussion.
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Single row or single column
With a single row, the top traversal consumes every cell and advances top beyond bottom. The later checks prevent a reverse traversal from appending that row again. With a single column, the downward traversal consumes the remaining cells and reduces right below left; the checks prevent the bottom and left traversals from repeating them.
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Simulate direction and turn when blocked
Another model tracks a current row, column, and direction. Begin at the upper-left moving right. After visiting a cell, inspect the next location: if it is outside the matrix or already visited, rotate clockwise; then continue. Stop after visiting exactly m*n cells.
A separate visited grid makes the turning rule explicit and leaves the input matrix unchanged. The cited LeetCode Wiki (Doocs) solution uses this approach and gives a complexity of O(mn) time and O(mn) auxiliary space; see 54. Spiral Matrix.
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Choose the model that suits the explanation
| Approach | Visual explanation | Extra memory | Input matrix | Stopping and edge-case logic |
|---|---|---|---|---|
| Shrinking boundaries | Shows the unvisited rectangle contracting after each edge. | Not quantified in the cited material. | Traverses the matrix without requiring value changes. | Continue while boundaries have not crossed; guard later edges against one-row and one-column leftovers. |
| Direction simulation with visited grid | Shows each move and clockwise turn when the next cell is blocked. | O(mn) auxiliary space in the cited solution. | Uses a separate visited grid rather than modifying matrix values. | Stop after exactly m*n visits; turn when the next cell is out of bounds or visited. |
For a visual walkthrough, shrinking boundaries make each perimeter and its inward movement easy to follow. For a move-by-move simulation, tracking direction and visited cells makes each turn explicit. Both produce the same clockwise order.
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- Confirm the output has
m*nvalues for the stated nonempty rectangular input. - Check the single-row and single-column cases, plus matrices with an odd number of rows or columns.
- For the boundary model, ensure every edge traversal stays within the current bounds and later edges are guarded after boundaries shrink.
- For direction simulation, rotate before moving into an out-of-bounds or previously visited cell.
A cited alternative temporarily adds 300 to visited matrix values. That marking trick depends on LeetCode 54’s stated value range, -100 <= matrix[i][j] <= 100, and changes the input during traversal. It is not a general-purpose substitute for a visited grid or boundary tracking. The problem’s stated dimensions are 1 <= m, n <= 10; see the problem constraints.
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