Algorithms and Data Structures – Spiral Matrix

--- Directions
Write a function that accepts an integer N
and returns a NxN spiral matrix.
--- Examples
matrix(2)
[[1, 2],
[4, 3]]
matrix(3)
[[1, 2, 3],
[8, 9, 4],
[7, 6, 5]]
matrix(4)
[[1, 2, 3, 4],
[12, 13, 14, 5],
[11, 16, 15, 6],
[10, 9, 8, 7]]
1. How to Approach
  • Create empty array of arrays called 'result'
  • Create a counter variable, starting at 1
  • As long as (start column <= end column) AND (start row <= end row)
    • Loop from start column to end column
      • At result[start_row][i] assign counter variable.
      • increment counter
    • Increment start row
    • Loop from start row to end row
      • At result [i][end_column] assign counter variable
      • Increment counter
    • Decrement end row
    • ... repeat for other two sides

2. Solution

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function matrix(n) {
  const results = [];
  for (let i = 0; i < n; i++) {
    results.push([]);
  }
  let counter = 1;
  let startColumn = 0;
  let endColumn = n - 1;
  let startRow = 0;
  let endRow = n - 1;
  while (startColumn <= endColumn && startRow <= endRow) {
    // Top row
    for (let i = startColumn; i <= endColumn; i++) {
      results[startRow][i] = counter;
      counter++;
    }
    startRow++;
    // Right column
    for (let i = startRow; i <= endRow; i++) {
      results[i][endColumn] = counter;
      counter++;
    }
    endColumn--;
    // Bottom row
    for (let i = endColumn; i >= startColumn; i--) {
      results[endRow][i] = counter;
      counter++;
    }
    endRow--;
    // start column
    for (let i = endRow; i >= startRow; i--) {
      results[i][startColumn] = counter;
      counter++;
    }
    startColumn++;
  }
  return results;
}
cs

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