前端手写题

目录

数组对比

获取当前元素（含）下所有元素出现次数 

 数组去重

数组降维

获取数组最大数

获取数组第二大数

获取数组中每项出现的次数

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数组对比

[1,2,3] ===[1,2,3] //false
JSON.stringify([1,2,3]) === JSON.stringify([1,2,3]) //true

获取当前元素（含）下所有元素出现次数 

<body>
  <div id="rich">
    <ul>
      <li><a>A</a></li>
      <li><a>B</a></li>
      <li><a>C</a></li>
    </ul>
    <h3>Hello</h3>
    <div>
      <input/>
      <button></button>
    </div>
  </div>
</body>
</html>
<script>
  let allElement = [document.querySelector('#rich'),...document.querySelector('#rich').getElementsByTagName("*")]
  console.log('元素展示：',allElement)
  let allNum = [...allElement].reduce((old,news)=>{
    let tagName = news.tagName
    if(old[tagName]){
      old[tagName]++
    }else{
      old[tagName] = 1
    }
    return old
  },{})
  console.log('最终展示：',allNum)
</script>

 结果展示 

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 数组去重

//1.使用reduce()方法去重
//注：使用reduce()至少得保证数组有2项才能执行；如果给了初始值，你至少得保证数组有一项才能执行。
    let arr = [1, 2, 2, 1, 4, null, null].reduce((accumulator, current) => {
      return accumulator.includes(current) ? accumulator : accumulator.concat(current)
    }, [])

2,使用new Set()去重
    let arr = [...new Set([1, 2, 2, 1, 4, null, null])]

3,多维去重 例如： [1,1,3,4,5,[2,1,12,[22,1,100]],32,[20,12,200]]
function deWeight(list,arrs){
    let newList = list.reduce((old,news)=>{
        if(!Array.isArray(news)){
            if(old.indexOf(news)===-1){
                old.push(news)
            }
        }else{
            return deWeight(news,old)
        }
    return old
    },arrs)
    return newList
}
deWeight(list,[])

数组降维

//1,多维数组降维flat会更简单，当然flat存在兼容问题：
//（Infinity参数代表无限深度，无参数默认降维一层）
let arr = [0,[1],[2, 3],[4, [5, 6, 7]]].flat(Infinity);// [0, 1, 2, 3, 4, 5, 6, 7]

//2，通过reduce，可以结合concat方法：
let arr = [[1,2],[3,4],[5,6]].reduce((accumulator, current)=>accumulator.concat(current),[]);//[1, 2, 3, 4, 5, 6]

//3，多维数组嵌套，reduce：
let arr = [0,[1],[2, 3],[4, [5, 6, 7]]];

let dimensionReduction = function (arr) {
    return arr.reduce((accumulator, current) => {
        return accumulator.concat(
            Array.isArray(current) ? 
            dimensionReduction(current) : 
            current
            );
    }, []);
}
dimensionReduction(arr); //[0, 1, 2, 3, 4, 5, 6, 7]

获取数组最大数

let arr = [1,2,5,34,34,43,21,8,6]
arr.reduce((old,news)=>{
    if(old['max']&&old['max']<news){
         old['max'] = news
    }else if(!old['max']){
        old['max'] = news
    }
    return old
},{}) //{max:43}

获取数组第二大数

let arr = [1,2,5,34,8,6,34,34,22,22,77,7,66,7,77,66]
arr.reduce((old,news,index,list)=>{
    if(index>=2){
        if(news>old['max2']){
            if(news>=old['max']){
                //为了区分存在相同数值，影响第二值的判断
                if(news!==old['max'])old['max2'] = old['max']
                old['max'] = news
            }else{
                old['max2']= news
            }
        }
        
    }else{
        if(list[0]>list[1]){
            old['max'] = list[0]
             old['max2'] = list[0]
        }else{
             old['max'] = list[1]
             old['max2'] = list[0]
        }
    }
    return old
},{}) //{max: 77, max2: 66}

获取数组中每项出现的次数

let arrs = [1,2,3,6,6,5,4,5,6,5,3,2]
arrs.reduce((old,news)=>{
    if(old[news]){
       old[news]++
    }else{
        old[news] = 1
    }
    return old
},{})