js数组去重

对于如下对象数组

[{id: 0, name: "name1"}, {id: 1, name: "name2"},{id: 1, name: "name2"},{id: 1, name: "name2"}, {id: 2, name: "name3"}, {id: 0, name: "name4"}]

现在想要将id相同的对象的name拼起来，笨的算法如下：

var ids = [];
        var ret = [];
        var map = {};
        var arr = [{id: 0, name: "name1"}, {id: 1, name: "name2"},{id: 1, name: "name2"},{id: 1, name: "name2"}, {id: 2, name: "name3"}, {id: 0, name: "name4"}];        //循环调用
        for (let i = 0; i < arr.length; i++) {
            map["" + arr[i].id] = arr[i];
        }
        var mulObj = {};
        var mulArr = [];
        for (var i = 0; i < arr.length; i++) {
            var obj = arr[i];   //当前对象
            if (ids.indexOf(obj.id) != -1) {          //如果包含，先取出该对象，然后遍历查找重复对象
                mulObj = map["" + obj.id];        //重复的对象+obj已有对象-----合并
                mulArr.push(mulObj);
            } else {
                ids.push(obj.id);
                ret.push(obj);
            }
        }
        for (let i = 0; i < ret.length; i++) {
            for(let j=0;j<mulArr.length;j++){
                if(ret[i].id == mulArr[j].id){      //说明重复
                    ret[i].name += mulArr[j].name;
                }
            }
        }
        console.log(ret);

对于形如这种字符串或数字数组去重：

[1,2,3,4,2,6,3,2,6,6,8]

简单算法实现如下：

let ret = [];
let hash = {};
for (let i = 0; i < arr.length; i++) {
    let item = arr[i];
    let key = typeof(item) + item;
    if (hash[key] !== 1) {
        ret.push(item);
        hash[key] = 1;
    }
}
return ret;