数组去重的方法

思路一:

  1. 双层循环,外层循环元素,内层循环时比较值

  2. 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-1; 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];
    console.log(arra.distinct()) //[3,4,2,1];

    思路二:利用splice直接在原数组进行操作

    1. 双层循环,外层循环元素,内层循环时比较值

    2. 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

    思路三: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

    思路四、数组递归去重

    1. 运用递归的思想

    2. 先排序,然后从最后开始比较,遇到相同,则删除

    Array.prototype.distinct = function (){
        var arr = this,
            len = arr.length;
    
        arr.sort(function(a,b){        //对数组进行排序才能方便比较
            return a - b;
        })
    
        function loop(index){
            if(index >= 1){
                if(arr[index] === arr[index-1]){
                    arr.splice(index,1);
                }
                loop(index - 1);    //递归loop函数进行去重
            }
        }
        loop(len-1);
    
        return arr;
    };
    
    var a = [1,2,3,4,5,6,5,3,2,4,56,4,1,2,1,1,1,1,1,1,56,45,56];
    var b = a.distinct();
    console.log(b.toString());        //1,2,3,4,5,6,45,56

    思路五、利用indexOf以及forEach

    Array.prototype.distinct = function (){
        var arr = this,
            result = [],
            len = arr.length;
    
        arr.forEach(function(v, i ,arr){        //这里利用map,filter方法也可以实现
            var bool =  arr.indexOf(v,i+1);        //从传入参数的下一个索引值开始寻找是否存在重复 v是数组中的每一个元素,i是数组中该元素对应的索引值
            if(bool === -1){
                result.push(v);
            }
        })
    
        return result;
    };
    
    var a = [1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,2,3,3,2,2,1,23,1,23,2,3,2,3,2,3];
    var b = a.distinct();
    console.log(b.toString());    //1,23,2,3
原文地址:https://www.cnblogs.com/pengwanxing/p/5518319.html