方法一:
双层循环,外层循环元素,内层循环时比较值
如果有相同的值则跳过,不相同则push进数组
Array.prototype.distinct = function(){
var arr = this,result = [], i,j,len = arr.length;
for(i = 0; i < len; i++){
for(j = i + 1; j < len; j++){
if(arr[i] === arr[j]){
j = ++i;
}
}
result.push(arr[i]);
}
return result;
}
var arra = [1,2,3,4,4,1,1,2,1,1,1];
arra.distinct(); //返回[3,4,2,1]
方法二:利用splice直接在原数组进行操作
双层循环,外层循环元素,内层循环时比较值
值相同时,则删去这个值
注意点:删除元素之后,需要将数组的长度也减1.
Array.prototype.distinct = function (){
var arr = this,i,j,len = arr.length;
for(i = 0; i < len; i++){
for(j = i + 1; j < len; j++){
if(arr[i] == arr[j]){
arr.splice(j,1);
len--;
j--;
}
}
}
return arr;
};
var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,];
var b = a.distinct();
console.log(b.toString()); //1,2,3,4,5,6,56
优点:简单易懂
缺点:占用内存高,速度慢
方法三:利用对象的属性不能相同的特点进行去重
Array.prototype.distinct = function (){
var arr = this,i,obj = {},result = [],len = arr.length;
for(i = 0; i< arr.length; i++){
if(!obj[arr[i]]){ //如果能查找到,证明数组元素重复了
obj[arr[i]] = 1;
result.push(arr[i]);
}
}
return result;
};
var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,];
var b = a.distinct();
console.log(b.toString()); //1,2,3,4,5,6,56
原文地址:https://www.cnblogs.com/NatChen/p/9818705.html