Spiral Matrix LeetCode

spiralMatrix.js

/*
Given an m x n matrix, return all elements of the matrix in spiral order.
 
Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
 
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
*/

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
 var spiralOrder = function(matrix) {
    if (matrix.length == 0) {
        return [];
    }
    let result = [];
    let rowStart = 0;
    let rowEnd = matrix.length - 1;
    let colStart = 0;
    let colEnd = matrix[0].length - 1;
    while (true) {
        // top
        for (let i = colStart; i <= colEnd; i++) {
            result.push(matrix[rowStart][i]);
        }
        rowStart++;
        if (rowStart > rowEnd) {
            return result;
        }
        // right
        for (let i = rowStart; i <= rowEnd; i++) {
            result.push(matrix[i][colEnd]);
        }
        colEnd--;
        if (colEnd < colStart) {
            return result;
        }
        // bottom
        for (let i = colEnd; i >= colStart; i--) {
            result.push(matrix[rowEnd][i]);
        }
        rowEnd--;
        if (rowEnd < rowStart) {
            return result;
        }
        // left
        for (let i = rowEnd; i >= rowStart; i--) {
            result.push(matrix[i][colStart]);
        }
        colStart++;
        if (colStart > colEnd) {
            return result;
        }
    }
        return result;
  };

  console.log(
    spiralOrder([[2, 3, 4], [5, 6, 7], [8, 9, 10], [11, 12, 13], [14, 15, 16]])
  );
  console.log(spiralOrder([[7], [9], [6]]));
  console.log(spiralOrder([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]));
  console.log(spiralOrder([[1, 2, 3], [4, 5, 6], [7, 8, 9]]));