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编程题:用最简洁代码实现 indexOf 方法 #58

Description

@sisterAn

indexOf 有两种:

String.prototype.indexOf()

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

str.indexOf(searchValue [, fromIndex])
// fromIndex 默认值为 0

Array.prototype.indexOf()

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

arr.indexOf(searchElement[, fromIndex])

解答

String.prototype.indexOf()

解题思路:正则,字符串匹配

function sIndexOf(str, searchStr, fromIndex = 0){
    var regex = new RegExp(`${searchStr}`, 'ig')
    regex.lastIndex = fromIndex
    var result = regex.exec(str)
    return result ? result.index : -1
}

// 测试
var paragraph = 'The quick brown fox jumps over the lazy dog. If the dog barked, was it really lazy?'
var searchTerm = 'dog'
// 测试一:不设置 fromIndex
console.log(sIndexOf(paragraph, searchTerm))
// 40
console.log(paragraph.indexOf(searchTerm));
// 40
// 测试二:设置 fromIndex
console.log(sIndexOf(paragraph, searchTerm, 41))
// 52
console.log(paragraph.indexOf(searchTerm, 41));
// 52

测试成功

Array.prototype.indexOf()

解题思路:遍历匹配

function aIndexOf(arr, elem, fromIndex = 0){
    if(!elem) return -1
    for(let i = fromIndex; i < arr.length; i++) {
        if(arr[i] === elem) return i
    }
    return -1
}

// 测试
var beasts = ['ant', 'bison', 'camel', 'duck', 'bison']
// 测试一:不设置 fromIndex
console.log(aIndexOf(beasts, 'bison'))
// 1
console.log(beasts.indexOf('bison'))
// 1
// 测试二:设置 fromIndex
console.log(aIndexOf(beasts, 'bison', 2))
// 4
console.log(beasts.indexOf('bison', 2))
// 4

测试成功

总结一下
function indexOf(items, item, fromIndex = 0) {
    let isArray = Array.isArray(items);
    let isString = Object.prototype.toString.call(items) == '[object String]';
    if (!isArray && !isString) throw new SyntaxError();
    if(isArray) return sIndexOf(items, item, fromIndex)
    else return aIndexOf(items, item, fromIndex)
}

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)

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