Leetcode - 54. Spiral Matrix

Javascript Code

/**
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function (matrix) {

    let top = 0, right = matrix[0].length, bottom = matrix.length, left = 0;

    let res = [];

    while (top < bottom && left < right) {

        // complete the top row ( right to left)
        for (let i = left; i < right; i++) {
            res.push(matrix[top][i]);
        }
        top++;

        // complete the right column ( top to bottom )
        for (let i = top; i < bottom; i++) {
            res.push(matrix[i][right - 1])
        }
        right--;

        // Check if there are still rows and columns left
        if (!(left < right && top < bottom)) break;


        // complete the bottom row (left to right)
        for (let i = right-1; i >= left; i--) {
            res.push(matrix[bottom-1][i])
        }
        bottom--;

        // complete the left column (bottom to top)
        for (let i = bottom - 1; i >= top; i--) {
            res.push(matrix[i][left]);
        }
        left++;
    }

    return res;

};