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https://gitee.com/coder-xiaomo/leetcode-problemset
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62 lines
13 KiB
JSON
62 lines
13 KiB
JSON
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"title": "Promise Pool",
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"content": "<p>Given an array of asyncronous functions <code>functions</code> and a <strong>pool limit</strong> <code>n</code>, return an asyncronous function <code>promisePool</code>. It should return a promise that resolves when all the input functions resolve.</p>\n\n<p><b>Pool limit</b> is defined as the maximum number promises that can be pending at once. <code>promisePool</code> should begin execution of as many functions as possible and continue executing new functions when old promises resolve. <code>promisePool</code> should execute <code>functions[i]</code> then <code>functions[i + 1]</code> then <code>functions[i + 2]</code>, etc. When the last promise resolves, <code>promisePool</code> should also resolve.</p>\n\n<p>For example, if <code>n = 1</code>, <code>promisePool</code> will execute one function at a time in series. However, if <code>n = 2</code>, it first executes two functions. When either of the two functions resolve, a 3rd function should be executed (if available), and so on until there are no functions left to execute.</p>\n\n<p>You can assume all <code>functions</code> never reject. It is acceptable for <code>promisePool</code> to return a promise that resolves any value.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nfunctions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 2\n<strong>Output:</strong> [[300,400,500],500]\n<strong>Explanation:</strong>\nThree functions are passed in. They sleep for 300ms, 400ms, and 200ms respectively.\nAt t=0, the first 2 functions are executed. The pool size limit of 2 is reached.\nAt t=300, the 1st function resolves, and the 3rd function is executed. Pool size is 2.\nAt t=400, the 2nd function resolves. There is nothing left to execute. Pool size is 1.\nAt t=500, the 3rd function resolves. Pool size is zero so the returned promise also resolves.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:\n</strong>functions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 5\n<strong>Output:</strong> [[300,400,200],400]\n<strong>Explanation:</strong>\nAt t=0, all 3 functions are executed. The pool limit of 5 is never met.\nAt t=200, the 3rd function resolves. Pool size is 2.\nAt t=300, the 1st function resolved. Pool size is 1.\nAt t=400, the 2nd function resolves. Pool size is 0, so the returned promise also resolves.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong>\nfunctions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 1\n<strong>Output:</strong> [[300,700,900],900]\n<strong>Explanation:</strong>\nAt t=0, the 1st function is executed. Pool size is 1.\nAt t=300, the 1st function resolves and the 2nd function is executed. Pool size is 1.\nAt t=700, the 2nd function resolves and the 3rd function is executed. Pool size is 1.\nAt t=900, the 3rd function resolves. Pool size is 0 so the returned promise resolves.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= functions.length <= 10</code></li>\n\t<li><code><font face=\"monospace\">1 <= n <= 10</font></code></li>\n</ul>\n",
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"translatedTitle": "Promice 对象池",
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"translatedContent": "<p>请你编写一个异步函数 <code>promisePool</code> ，它接收一个异步函数数组 <code>functions</code> 和 <strong>池限制</strong> <code>n</code>。它应该返回一个 promise 对象，当所有输入函数都执行完毕后，promise 对象就执行完毕。</p>\n\n<p><strong>池限制</strong> 定义是一次可以挂起的最多 promise 对象的数量。<code>promisePool</code> 应该开始执行尽可能多的函数，并在旧的 promise 执行完毕后继续执行新函数。<code>promisePool</code> 应该先执行 <code>functions[i]</code>，再执行 <code>functions[i + 1]</code>，然后执行 <code>functions[i + 2]</code>，等等。当最后一个 promise 执行完毕时，<code>promisePool</code> 也应该执行完毕。</p>\n\n<p>例如，如果 <code>n = 1</code> , <code>promisePool</code> 在序列中每次执行一个函数。然而，如果 <code>n = 2</code> ，它首先执行两个函数。当两个函数中的任何一个执行完毕后，再执行第三个函数(如果它是可用的)，依此类推，直到没有函数要执行为止。</p>\n\n<p>你可以假设所有的 <code>functions</code> 都不会被拒绝。对于 <code>promisePool</code> 来说，返回一个可以解析任何值的 promise 都是可以接受的。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>\nfunctions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 2\n<b>输出：</b>[[300,400,500],500]\n<strong>解释</strong>\n传递了三个函数。它们的睡眠时间分别为 300ms、 400ms 和 200ms。\n在 t=0 时，执行前两个函数。池大小限制达到 2。\n当 t=300 时，第一个函数执行完毕后，执行第3个函数。池大小为 2。\n在 t=400 时，第二个函数执行完毕后。没有什么可执行的了。池大小为 1。\n在 t=500 时，第三个函数执行完毕后。池大小为 0，因此返回的 promise 也执行完成。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：\n</strong>functions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 5\n<b>输出：</b>[[300,400,200],400]\n<strong>解释：</strong>\n在 t=0 时，所有3个函数都被执行。池的限制大小 5 永远不会满足。\n在 t=200 时，第三个函数执行完毕后。池大小为 2。\n在 t=300 时，第一个函数执行完毕后。池大小为 1。\n在 t=400 时，第二个函数执行完毕后。池大小为 0，因此返回的 promise 也执行完成。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nfunctions = [\n () => new Promise(res => setTimeout(res, 300)),\n () => new Promise(res => setTimeout(res, 400)),\n () => new Promise(res => setTimeout(res, 200))\n]\nn = 1\n<b>输出：</b>[[300,700,900],900]\n<strong>解释：</strong>\n在 t=0 时，执行第一个函数。池大小为1。\n当 t=300 时，第一个函数执行完毕后，执行第二个函数。池大小为 1。\n当 t=700 时，第二个函数执行完毕后，执行第三个函数。池大小为 1。\n在 t=900 时，第三个函数执行完毕后。池大小为 0，因此返回的 Promise 也执行完成。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 <= functions.length <= 10</code></li>\n\t<li><code><font face=\"monospace\">1 <= n <= 10</font></code></li>\n</ul>\n",
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"Initially execute all the functions until the queue fills up.",
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"Every time a function resolves, add a new promise to the queue if possible."
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