爱在桂子山：请问是线上还是线下呀？

小强不强：现在有笔试了，我是在校生想投实习的

鲮鲤猫：你好～上面事件循环输出题的结果应该是 1 2 4 timerStart timerEnd success，setTimeout对应的那个宏任务执行完毕才会交还线程，检查微任务队列是否有任务，若没有再从宏任务中取一个任务压入执行栈执行

牛客468429517号：加上回溯，就可以100%通过测试用例了， let n = readline().split(" ").map(Number), maze = []; //读数据 while ((line = readline())) maze.push(line.split(" ").map(Number)); let v = [...Array(n[0])].map(() => Array(n[1]).fill(0)); function find(i, j, road) { road.push([i, j]); //先推入0,0 maze[i][j] = 1; if (i == n[0] - 1 && j == n[1] - 1) { for (let i of road) print(`(${i[0]},${i[1]})`); } if (j + 1 < n[1] && maze[i][j + 1] == 0 && v[i][j + 1] == 0) { v[i][j + 1] = 1; find(i, j + 1, road); v[i][j + 1] = 0; road.pop(); } //右 if (i + 1 < n[0] && maze[i + 1][j] == 0 && v[i + 1][j] == 0) { v[i + 1][j] = 1; find(i + 1, j, road); v[i + 1][j] = 0; road.pop(); } //下 if (j - 1 >= 0 && maze[i][j - 1] == 0 && v[i][j - 1] == 0) { v[i][j - 1] = 1; find(i, j - 1, road); v[i][j - 1] = 0; road.pop(); } //左 if (i - 1 >= 0 && maze[i - 1][j] == 0 && v[i - 1][j] == 0) { v[i - 1][j] = 1; find(i - 1, j, road); v[i - 1][j] = 0; road.pop(); } //上 } find(0, 0, []);