LeetCode 54. Spiral Matrix (javascript)

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

Solution:

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function(matrix) {
    if (matrix === null) return null
    let res = [];
    let l = 0;
    let t = 0;
    let r = matrix[0].length - 1;
    let b = matrix.length - 1;
    const total = (r + 1) * (b + 1);
    let d = 0;
    let x = 0;
    let y = 0;
    while (res.length < total - 1) {
        // right
        if (d === 0) {
            while (x < r) res.push(matrix[y][x++]);
            t++; // reach end, turn 90 degree, move down
        // down 
        } else if (d === 1) {
            while (y < b) res.push(matrix[y++][x]);
            r--; // reach end, turn 90 degree, move left
        // left
        } else if (d === 2) {
            while (x > l) res.push(matrix[y][x--]);
            b--; // reach end, turn 90 degree, move up
        // up
        } else if (d === 3) {
            while (y > t) res.push(matrix[y--][x]);
            l++; // reach end, turn 90 degree, move up right
        }
        d = (d + 1) % 4;
    }
    // last move to the center if exist
    if (res.length !== total) 
        res.push(matrix[y][x]);
    
    return res;
};