Given an m x n matrix, return all elements of the matrix in spiral order.

spiralMatrix.js

/**
 * @param {number[][]} matrix
 * @return {number[]}
 * https://underdash.web.app
 * https://leetcode.com/problems/spiral-matrix/
 * https://underdash.tech/spiral-matrix/
 
 * Copyright reserved by Taha Shieenavaz (Underdash)
 */

const spiralOrder = matrix => {
    const { abs } = Math
    const rows = matrix.length, cols = matrix[0].length
    const map = Array(rows).fill().map(row => {
        return Array(cols).fill(false)
    })
    const result = []
    const direction = { x: 1, y: 0 }
    let i = 0, j = 0
    
    while( map.flat().filter( a => !a ).length > 0 ) {
        if(
            ( direction.x == -1 && i == 0 )       ||
            ( direction.y == -1 && j == 0 )       ||
            ( direction.x == 1 && i == cols - 1 ) ||
            ( direction.y == 1 && j == rows - 1 ) ||
            map[j+direction.y][i+direction.x]
        ){
            if( abs(direction.x) ) {
                direction.y = direction.x
                direction.x = 0
            }else if( abs(direction.y) ) {
                direction.x = -1 * direction.y
                direction.y = 0
            }
        }
        
        result.push(matrix[j][i])
        map[j][i] = true
        i += direction.x
        j += direction.y
    }
    return result
};

Read more about this solution: https://underdash.pro/spiral-matrix/