需要模拟的题目没有啥算法技巧，只能模拟其变化的轨迹，得到答案

54.螺旋矩阵

按照「形状」进行模拟

/**
 * 按形状模拟
 * @param {number[][]} matrix
 * @return {number[]}
 */
var spiralOrder = function (matrix) {
    let ans = new Array();
    let m = matrix.length, n = matrix[0].length;
    circle(matrix, 0, 0, m - 1, n - 1);
    return ans;
    function circle(matrix, x1, y1, x2, y2) {
        if (x2 < x1 || y2 < y1) return;
        // 只有一行时，按「行」遍历
        if (x1 === x2) {
            for (let i = y1; i <= y2; i++) {
                ans.push(matrix[x1][i]);
            }
            return;
        }
        // 只有一列时，按「列」遍历
        if (y1 === y2) {
            for (let i = x1; i <= x2; i++) {
                ans.push(matrix[i][y2]);
            }
            return;
        }
        // 圈
        for (let i = y1; i < y2; i++)
            ans.push(matrix[x1][i]);
        for (let i = x1; i < x2; i++)
            ans.push(matrix[i][y2]);
        for (let i = y2; i > y1; i--)
            ans.push(matrix[x2][i]);
        for (let i = x2; i > x1; i--)
            ans.push(matrix[i][y1]);
        // 往里收一圈，继续遍历
        circle(matrix, x1 + 1, y1 + 1, x2 - 1, y2 - 1);
    }
};

按照「方向」进行模拟

/**
 * 按方向模拟
 * @param {number[][]} matrix
 * @return {number[]}
 */
const direction = [
    [0, 1], [1, 0], [0, -1], [-1, 0]
];
var spiralOrder = function (matrix) {
    let ans = new Array();
    let m = matrix.length, n = matrix[0].length;
    for (let x = 0, y = 0, d = 0, i = 0; i < n * m; i++) {
        ans.push(matrix[x][y]);
        matrix[x][y] = Infinity;
        let nx = x + direction[d][0];
        let ny = y + direction[d][1];
        if (nx < 0 || nx >= m || ny < 0 || ny >= n || matrix[nx][ny] === Infinity) {
            d = (d + 1) % 4;
            nx = x + direction[d][0];
            ny = y + direction[d][1];
        }
        x = nx;
        y = ny;
    }
    return ans;
};