题解 | 输入n个整数，输出其中最小的k个

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void async function () {
    // Write your code here
    let a = await readline()
    let line = a.split(' ')
    let b = await readline();
    let array = b.split(' ').map(a => parseInt(a)).sort((a,b)=>a-b);
    let num = parseInt(line[1])
    //console.log(num)
    let ans = ""
    for(let i=0;i<num;i++){
        ans +=array[i] +" "
    }
    console.log(ans.trim())
}()

写得乱七八糟的，顺便要记得readline返回的是一个Promise，解析后才是字符串，因为Promise没有split()方法所以不能直接对await readline()后面接上split()等Promise没有的方法