spiral-matrix.js

// Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
//
// Example 1:
//
//
// Input:
// [
//  [ 1, 2, 3 ],
//  [ 4, 5, 6 ],
//  [ 7, 8, 9 ]
// ]
// Output: [1,2,3,6,9,8,7,4,5]
//
//
// Example 2:
//
// Input:
// [
//   [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]
//


/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function(matrix) {
    var dir = 2; //0: left, 1: up, 2: right: 3: down
    var m = matrix.length,
        n = (matrix[0] || []).length;
    var mark = new Array(m * n);
    var result = new Array(m * n);
    var y = 0, x = 0;
    var index = 0;
    var next = function() {
        var newX, newY;
        while (true) {
            newX = x, newY = y;
            switch (dir) {
                case 0: newX = x - 1; break;
                case 1: newY = y - 1; break;
                case 2: newX = x + 1; break;
                case 3: newY = y + 1; break;
                default: break;
            }
            if (newX < 0 || newX >= n || newY < 0 || newY >= m || mark[newY * n + newX]) {
                dir = (dir + 1) % 4;
            } else {
                break;
            }
        }
        x = newX;
        y = newY;
    };
    
    while (index < m * n) {
        result[index] = matrix[y][x];
        mark[y * n + x] = true;
        if (index !== m*n-1) {
            next();
        }
        ++index;
    }
   
    return result;
};