Spiral Matrix: the idea to remember.

The core insight

Traverse the top row, right column, bottom row and left column, shrinking the corresponding boundary after each side.

Why it works

Each pass consumes exactly the current outer layer. The remaining rectangle is the same smaller problem.

The mistake to avoid

Check bounds before the bottom and left sides so a single remaining row or column is not visited twice.

Commented Java solution

Use java.util.* imports and a surrounding solution class. Tree snippets assume the standard TreeNode definition; linked lists use ListNode and Clone Graph uses a Node with val and neighbors; Copy Random List uses Node with val, next and random. Read the input assumptions above.

List<Integer> spiralOrder(int[][] a) {
    List<Integer> out = new ArrayList<>();
    if (a.length == 0) return out;
    int top = 0, bottom = a.length - 1, left = 0, right = a[0].length - 1;
    while (top <= bottom && left <= right) {
        for (int c = left; c <= right; c++) out.add(a[top][c]); top++;
        for (int r = top; r <= bottom; r++) out.add(a[r][right]); right--;
        if (top <= bottom) { for (int c = right; c >= left; c--) out.add(a[bottom][c]); bottom--; }
        if (left <= right) { for (int r = bottom; r >= top; r--) out.add(a[r][left]); left++; }
    }
    return out;
}

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