如下
arr:[12,12,17,10,12,10]
并且超过三个还得删除一个让他始终保持两个
建议学一些简单的方法,有助于提升效率
三行代码可以搞定,如果谁更少,欢迎交流
var arr = [12,12,17,10,12,10];
const map = new Map()
arr = arr.filter(item => !map.has(item) && map.set(item, 1))
arr.length = arr.length >= 2 ? 2 : arr.length
```javascript
//方法一
var arr = [12,12,17,10,12,10];
var temp = [];
for(var i=0;i<arr.length;i++){
if(temp.indexOf(arr[i]) == -1){
temp.push(arr[i]);
}
}
arr = temp;
console.log(arr);
//方法二
var arr = [12,12,17,10,12,10];
var obj = {};
for(var i=0;i<arr.length;i++){
if(!obj[arr[i]]){
obj[arr[i]] = 1;
}else{
obj[arr[i]]++;
}
}
arr = [];
for(var key in obj){
if(obj[key]>2){
arr.push(key);
}
}
console.log(arr);
```
具体实现如下:
function removeDuplicates(arr) {
var counts = {};
for (var i = 0; i < arr.length; i++) {
var num = arr[i];
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
for (var num in counts) {
if (counts[num] > 2) {
for (var i = 0; i < counts[num] - 2; i++) {
var index = arr.indexOf(parseInt(num));
arr.splice(index, 1);
}
}
}
return arr;
}
console.log(removeDuplicates([12,12,17,10,12,10]));