Array of integers obtained by spiralling outward anti-clockwise from the r and c, starting upward

Recently, I found a great resource of toy problems on GitHub and wanted to take a stab at the following problem.

As a starting base to wrap my head around the problem and to see if I could spot any pattern, I used the whiteboard. What I found were two things.

Write a function that accepts four arguments. The first two arguments are the size of the grid (h x w), filled with ascending integers from left to right, top to bottom, starting from 1. The next two arguments are the starting positions, the row (r) and column (c).

f(5, 5, 3, 3) // [ 13, 8, 7, 12, 17, 18, 19, 14, 9, 4, 3, 2, 1, 6, 11, 16, 21, 22, 23, 24, 25, 20, 15, 10, 5 ]

f(2, 4, 1, 2) // [ 2, 1, 5, 6, 7, 3, 8, 4 ]

Return an array of integers obtained by spiralling outward anti-clockwise from the r and c, starting upward.

var spiral = function (h, w, r, c) {
  // value to start the board
  var begin = 1;
  var board = [];
  // build the board
  for (var i = 0; i < h; i++) {
    var row = [];
    while (row.length < w) { row.push(begin++); }
    board.push(row);
  }

  // run length for each direction
  var run = 1;
  // normalize the indexes
  var row = r - 1, col = c - 1;

  var positions = [];
  var recurse = function () {
    // if the length is equal, then we have visited all possibilities
    if (positions.length === h * w) { return; }
    // direction up
    for (var i = 0; i < run; i++) {
      if (board[row] && board[row][col]) { positions.push(board[row][col]); }
      row--;
    }
    // direction left
    if (positions.length === h * w) { return; }
    for (var i = 0; i < run; i++) {
      if (board[row] && board[row][col]) { positions.push(board[row][col]); }
      col--;
    }
    run++;
    // direction down
    if (positions.length === h * w) { return; }
    for (var i = 0; i < run; i++) {
      if (board[row] && board[row][col]) { positions.push(board[row][col]); }
      row++;
    }
    // direction right
    if (positions.length === h * w) { return; }
    for (var i = 0; i < run; i++) {
      if (board[row] && board[row][col]) { positions.push(board[row][col]); }
      col++;
    }
    run++;

    recurse();
  };

  recurse();
  return positions;
};

Written on February 27, 2015