对象或者数组相等

实现一个方法，只要内容形态一致，则认为数组或对象相等

let obj1 = {a:[1,'2'],b:2};
let obj2 = {b:2,a:[1,'2']};
let arr1 = [1,2,{a:{c:1},b:2},'miya'];
let arr2 = [1,2,{b:2,a:{c:1}},"miya"];
var obj3 = {a:[1,'2',{name:'miya',age:18}],b:{c:[3,1,2]}};
var obj4 = {b:{c:[3,1,2]},a:[1,'2',{age:18,name:'miya'}]};
//比如上面的
obj1 === obj2
arr1 === arr2
obj3 === obj4

答案：

function eql01(obj1,obj2){
    if(Object.prototype.toString.call(obj1) !== Object.prototype.toString.call(obj2)){
        return false;
    }
    let sortObj = (r1,r2)=>{
        let obj = {}
        for(var i in r1){
            if(!r2[i]) return false;
            obj[i] = r2[i];
        }
        Object.assign(obj,r2);
        return obj;
    }
    if(Array.isArray(obj1)){
        obj1.map((val,key)=>{
         if(Object.prototype.toString.call(val) === "[object Object]"){
            let ss = sortObj(val,obj2[key]);
            obj2[key] = ss;            
         }
      })
    }else{
        let gg = sortObj(obj1,obj2);
        obj2 = gg;
    }    
    return JSON.stringify(obj1) === JSON.stringify(obj2);
}

eql02应该是最终版本了

function eql02(obj1,obj2){
    if(Object.prototype.toString.call(obj1) !== Object.prototype.toString.call(obj2)){
        return false;
    }
    if(Array.isArray(obj1)){
        obj1.map((val,key)=>{
            if(Object.prototype.toString.call(val) === "[object Object]"){
                    obj2[key] = Object.assign(val,obj2[key]);
            }
        })
    }else{
        obj2 = Object.assign(obj1,obj2);
    }
    return JSON.stringify(obj1) === JSON.stringify(obj2);
}

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