1.利用对象

思路：定义一个对象和数组，用对象来判断数组里的每一个元素是否一样，把数组的每一个都作为对象的key如果对象里面存在这个key当前就不添加到新数组中，没过没有就添加到新数组中

 const arry=[1,5,3,6,3,5,3,6,6,3,5,8,9,7,8,2,1,5]
  function quchong(arry){
    let obj={}
    let newarry=[]
    for(let a of arry){
       if(!obj[a]){
         obj[a]=1
         newarry.push(a)
       }
    }
    return newarry
  }
console.log(quchong(arry));

2.利用indexof

indexOf判断新数组是否有去重的数组的元素，没有的话就push进去

<script>
  const arry=[1,5,3,6,3,5,3,6,6,3,5,8,9,7,8,2,1,5]
  function quchong(arry){
    let newarry=[]
    for(let a of arry){
       if(newarry.indexOf(a)===-1){
         newarry.push(a)
       }
    }
    return newarry
  }
console.log(quchong(arry));
</script>

3.利用includes来判断重复

同一种思想 数组.includes(元素) 返回值是布尔值

<script>
  const arry=[1,5,3,6,3,5,3,6,6,3,5,8,9,7,8,2,1,5]
  function quchong(arry){
    let newarry=[]
    for(let a of arry){
       if(!newarry.includes(a)){
         newarry.push(a)
       }
    }
    return newarry
  }
console.log(quchong(arry));
</script>

4.filter和includes

<script>
  const arry=[1,5,3,6,3,5,3,6,6,3,5,8,9,7,8,2,1,5]
  function quchong(arry){
    let newarry=[]
     arry.filter(a=>{
          if(!newarry.includes(a)){
            newarry.push(a)
          }
       })
    return newarry
  }
console.log(quchong(arry));
</script>

5.forEach和includes

  const arry=[1,5,3,6,3,5,3,6,6,3,5,8,9,7,8,2,1,5]
  function quchong(arry){
    let newarry=[]
     arry.forEach(a=>{
          if(!newarry.includes(a)){
            newarry.push(a)
          }
       })
    return newarry
  }
console.log(quchong(arry));