René's URL Explorer Experiment

{"@context":"https://schema.org","@type":"DiscussionForumPosting","headline":"编程题：用最简洁代码实现 indexOf 方法","articleBody":"`indexOf` 有两种：\r\n\r\n\u003e String.prototype.indexOf() \r\n\r\n返回从 `fromIndex` 处开始搜索第一次出现的指定值的索引，如果未找到，返回 `-1`\r\n\r\n```js\r\nstr.indexOf(searchValue [, fromIndex])\r\n// fromIndex 默认值为 0\r\n```\r\n\r\n\r\n\r\n\u003e Array.prototype.indexOf()\r\n\r\n返回在数组中可以找到一个给定元素的第一个索引，如果不存在，则返回 `-1`\r\n\r\n```js\r\narr.indexOf(searchElement[, fromIndex])\r\n```\r\n\r\n#### 解答\r\n\r\n##### String.prototype.indexOf() \r\n\r\n**解题思路：正则，字符串匹配**\r\n\r\n```js\r\nfunction sIndexOf(str, searchStr, fromIndex = 0){\r\n    var regex = new RegExp(`${searchStr}`, 'ig')\r\n    regex.lastIndex = fromIndex\r\n    var result = regex.exec(str)\r\n    return result ? result.index : -1\r\n}\r\n\r\n// 测试\r\nvar paragraph = 'The quick brown fox jumps over the lazy dog. If the dog barked, was it really lazy?'\r\nvar searchTerm = 'dog'\r\n// 测试一：不设置 fromIndex\r\nconsole.log(sIndexOf(paragraph, searchTerm))\r\n// 40\r\nconsole.log(paragraph.indexOf(searchTerm));\r\n// 40\r\n// 测试二：设置 fromIndex\r\nconsole.log(sIndexOf(paragraph, searchTerm, 41))\r\n// 52\r\nconsole.log(paragraph.indexOf(searchTerm, 41));\r\n// 52\r\n```\r\n\r\n测试成功\r\n\r\n##### Array.prototype.indexOf()\r\n\r\n**解题思路：遍历匹配**\r\n\r\n```js\r\nfunction aIndexOf(arr, elem, fromIndex = 0){\r\n    if(!elem) return -1\r\n    for(let i = fromIndex; i \u003c arr.length; i++) {\r\n        if(arr[i] === elem) return i\r\n    }\r\n    return -1\r\n}\r\n\r\n// 测试\r\nvar beasts = ['ant', 'bison', 'camel', 'duck', 'bison']\r\n// 测试一：不设置 fromIndex\r\nconsole.log(aIndexOf(beasts, 'bison'))\r\n// 1\r\nconsole.log(beasts.indexOf('bison'))\r\n// 1\r\n// 测试二：设置 fromIndex\r\nconsole.log(aIndexOf(beasts, 'bison', 2))\r\n// 4\r\nconsole.log(beasts.indexOf('bison', 2))\r\n// 4\r\n```\r\n\r\n测试成功\r\n\r\n##### 总结一下\r\n\r\n```js\r\nfunction indexOf(items, item, fromIndex = 0) {\r\n    let isArray = Array.isArray(items);\r\n    let isString = Object.prototype.toString.call(items) == '[object String]';\r\n    if (!isArray \u0026\u0026 !isString) throw new SyntaxError();\r\n    if(isArray) return sIndexOf(items, item, fromIndex)\r\n    else return aIndexOf(items, item, fromIndex)\r\n}\r\n```\r\n\r\n你也可以尝试使用遍历匹配法解决 `sIndexOf` 问题（正则更简洁），这里不做介绍（和 `aIndexOf ` 差不多的套路，不同的是，`String` 类型可以一次匹配多个字符）","author":{"url":"https://github.com/sisterAn","@type":"Person","name":"sisterAn"},"datePublished":"2020-06-02T00:17:53.000Z","interactionStatistic":{"@type":"InteractionCounter","interactionType":"https://schema.org/CommentAction","userInteractionCount":8},"url":"https://github.com/58/JavaScript-Algorithms/issues/58"}