[Leetcode] JS 30 Sleep

Question

Given a positive integer millis, write an asynchronous function that sleeps for millis milliseconds. It can resolve any value.

Example 1:

Input: millis = 100
Output: 100
Explanation: It should return a promise that resolves after 100ms.
let t = Date.now();
sleep(100).then(() => {
  console.log(Date.now() - t); // 100
});

Example 2:

Input: millis = 200
Output: 200
Explanation: It should return a promise that resolves after 200ms.

Constraints:

• 1 <= millis <= 1000

My Solution

/**
 * @param {number} millis
 * @return {Promise}
 */
async function sleep(millis) {
    return new Promise(resolve => {setTimeout(resolve,millis)});
}

/** 
 * let t = Date.now()
 * sleep(100).then(() => console.log(Date.now() - t)) // 100
 */

Key Takeaways

The key rule in this question is that async functions always return a Promise. Since we see Promise.then() being used in the test cases, sleep() must return a Promise. We do not need to use await inside the sleep() function because we are not executing the Promise immediately, but rather returning a Promise that will be handled by .then().

Moreover, based on our last article about Promise here https://404-peace-not-found.ghost.io/ghost/#/site, you can also restructure your code with async/await instead of .then() .

First method

function sleep(millis) {
    return new Promise(resolve => setTimeout(resolve, millis));
}

async function test() {
    let t = Date.now();
    await sleep(100); // will wait Promise to be solved
    console.log(Date.now() - t); // about 100
}

test();

Second Method

async function sleep(millis) {
    let t = Date.now();
    await new Promise(resolve => { 
        setTimeout(() => { 
            console.log(Date.now() - t);
            resolve(); // we have to add resolve() here otherwise this await will get stuck
        }, millis); 
    });
}