某饭店要开发一个排队软件。非VIP用户需要排队，先到先得。V

class Customer{
	constructor(flag,time,number){
		this.vip=flag;
		this.time=time;
		this.phoneNumber=number;
	}
	getVip(){
		return this.vip;
	}
	getTime(){
		return this.time;
	}
	getNumber(){
		return this.phoneNumber;
	}
}
class Queue{
	constructor(){
		this.Q=[];
		this.vipQ=[];
		this.notVipQ=[];
	}
	add(c){
		if (c.vip) { //c是Customer的实例
			this.vipQ.push(c.phoneNumber);
		}else{
			this.notVipQ.push(c.phoneNumber);
		}
	}
	getCustomer(){
		if (!this.Q.length&&this.notVipQ.length) {
			if (!this.vipQ.length) {
				this.Q.push(this.notVipQ.shift());
			}else{
				this.Q.push(this.vipQ.shift());
			}
		}
	}
}
不晓得还要用current实现什么功能，还请各位大神指点！

	class queue:



	    def __init__(self,vip,nor):



	        if(type(vip)==list and type(nor)==list):



	            self.vip = vip



	            self.nor = nor



	        else :print 'vip and nor must be list'



	





	    def addcustom(self,name,phonenumber,ifvip):



	        if (ifvip):



	            self.vip.append([name,phonenumber])



	        else :



	            self.nor.append([name,phonenumber])



	





	    def next(self):



	        if (self.vip):



	            n = self.vip[0][0]



	            del(self.vip[0])



	            print 'next is ' + n



	        else :



	            n = self.nor[0][0]



	            del(self.nor[0])



	            print 'next is ' + n



	





	    def current(self):



	        str1 = '现在队列为：'



	        for i in self.vip:



	            str1 += i[0]



	            str1 += ', '



	        for i in self.nor:



	            str1 += i[0]



	            str1 +=', '



	        print str1



	



	用Python试着写了下，不知道算不算满足题目要求……