Question

我的nbSandwichs（int type）方法有问题它应该通过双重链表并计算相同类型的三明治出现的次数，一切都很好，除了打印0的最后一个三明治，这是我不明白的，我的检查方法是说它不存在但是当我创建一个get get方法时，它确实存在。我的nbSandwichs方法缺少什么条件？我的while循环实际上没有到达最后一个节点吗？

谢谢

main class : 
    Sandwich s1 = new Sandwich(1);
        Sandwich s1 = new Sandwich(1);
        Sandwich s2 = new Sandwich(15);
        Sandwich s3 = new Sandwich(15);
        Sandwich s4 = new Sandwich(4);
        Sandwich s5 = new Sandwich(15);

        APreparer a1 = new APreparer();
        a1.addfirst(s1); 
        a1.addfirst(s2);
        a1.addfirst(s3);
        a1.addfirst(s4);
        a1.addfirst(s5);

        System.out.println(a1.nbSandwichs(15)); // PRINTS : 3 OK 
        System.out.println(a1.nbSandwichs(1)); // PRINTS : 0 NOT OK 


    public class Sandwich {

        private int type;

        public Sandwich(int type) {
            this.type = type;
            commandes[type]++;
        }

    public class APreparer {

        private UneCommande first;
        private UneCommande last;

        public void addfirst(Sandwich sandwich) {
            UneCommande nouvelle = new UneCommande(sandwich);
            if (first == null) {
                first = nouvelle;
                last = nouvelle;

            } else {
                first = first.addFirst(sandwich);
            }
        }

    int nbSandwichs(int type) {
        if (first == null) {
            return 0;
        } else {
            return first.nbSandwichs(type);
        }
    }

    }


    public class UneCommande {

        private Sandwich sandwich;
        private UneCommande next;
        private UneCommande previous;

        public UneCommande(Sandwich sandwich) {
            this.sandwich = sandwich;
        }

        public UneCommande addFirst(Sandwich sandwich) {
            UneCommande current = this;
            UneCommande newSand = new UneCommande(sandwich);
            newSand.next = current;
            this.previous = newSand;

            return newSand;
        }
int nbSandwichs(int type) {
        int counter = 0;
        UneCommande current = this;

        if (!(check(type))) {
            return 0;
        } else {
            while (current.next != null) {
                if (current.sandwich.getType() == type) {
                    counter++;
                }
                current = current.next;
            }
        }
        return counter;
    }

    boolean check(int type) {
        UneCommande current = this;
        while (current != null) {
            if (current.sandwich.getType() == type) {
                System.out.println("EXIST");
                return true;
            }
            current = current.next;
        }

        return false;
    }
}

Answer 1

您的循环计算节点的时间与current.next != null一样长。当current是列表中的最后一个节点时，current.next将为null，因此不会计算在内。

计算其中有多少＆＃34;元素＆＃34;出现在双向链表中

1 个答案: