René's URL Explorer Experiment

{"@context":"https://schema.org","@type":"DiscussionForumPosting","headline":"腾讯：不产生新数组，删除数组里的重复元素","articleBody":"数组去重的方式有很多，我们可以使用 Set 去重、filter 过滤等，详见 [携程\u0026蘑菇街\u0026bilibili：手写数组去重、扁平化函数](https://mp.weixin.qq.com/s/NGl9nBkIruuzYkxqyPnB1A) ，但三种解法（Set、filter、reducer）都产生了新数组：\r\n\r\n\u003e MDN ： `filter()` 方法创建一个新数组, 其包含通过所提供函数实现的测试的所有元素。\r\n\r\n那么我们如何在不产生性数组的情况下删除数组中的重复元素喃？\r\n\r\n### 方式一：排序去重\r\n\r\n\u003e MDN：`sort()` 方法用[原地算法](https://en.wikipedia.org/wiki/In-place_algorithm)对数组的元素进行排序，并返回数组。默认排序顺序是在将元素转换为字符串，然后比较它们的UTF-16代码单元值序列时构建的\r\n\r\n```js\r\nconst removeDuplicates = (nums) =\u003e {\r\n    // 原地排序\r\n    nums.sort()\r\n    // 去重\r\n    let len = 1\r\n    for (let i = 1; i \u003c nums.length; i++)\r\n        if (nums[i] != nums[i-1]) nums[len++] = nums[i];\r\n    // 删除重复项\r\n    nums.splice(len)\r\n    return nums\r\n}\r\n\r\n// 测试\r\nremoveDuplicates([1, 2, 3, 1, 3])\r\n// [1, 2, 3]\r\n```\r\n\r\n### 方式二：优化\r\n\r\n```js\r\nconst removeDuplicates = (nums) =\u003e {\r\n    let len = nums.length - 1\r\n    for(let i = len; i\u003e=0; i--) {\r\n        if(nums.indexOf(nums[i]) != i) {\r\n            nums[i] = nums[len --]\r\n        }\r\n    }\r\n    // 删除重复项\r\n    nums.splice(len+1)\r\n    return nums\r\n}\r\n// 测试\r\nremoveDuplicates([1, 2, 3, 1, 3])\r\n// [1, 2, 3]\r\n```","author":{"url":"https://github.com/sisterAn","@type":"Person","name":"sisterAn"},"datePublished":"2020-12-06T23:50:07.000Z","interactionStatistic":{"@type":"InteractionCounter","interactionType":"https://schema.org/CommentAction","userInteractionCount":7},"url":"https://github.com/135/JavaScript-Algorithms/issues/135"}