2.3

Topic: Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node.

EXAMPLE Input: the node c from the linked list a->b->c->d->d->e

    Output: nothing is returned, but the new linked list looks like a->b->d->e

// 关键：考虑要删的节点是head, in the middle, last, empty list.

           (1)(2)头和中间节点适用，(3)尾节点:将数据设为某特殊字符，打印时不打印它，(4)空/只有它:直接返回。

// 方法：由于找不到此节点的上一个节点，所以把下一个节点的数据给要删的，下一个的next给这一个的next.实际删了下一个节点,但保留了下一个节点的数据，覆盖了这个节点的数据。（Cannot access the head of the linked list, copy the data from next node over to the current node, delete the next node）

public class List { int data;    List next;    public List(int d) {        this.data = d;        this.next = null;    }        void appendToTail(int d) {//依次在最后一个节点的后面追加元素        List end = new List(d);        List n = this;        while (n.next != null) {//判断是否是最后一个节点，如果不是，往后移            n = n.next;        }        n.next = end;//把这个节点设为最后一个节点    }        void print() {        List n = this;        System.out.print("{");        while (n != null) {            if (n.next != null)                System.out.print(n.data + ", ");            else                System.out.println(n.data + "}");            n = n.next;        }    }            public static boolean deleteNode(List n){        if(n==null||n.next==null) return false;        n.data=n.next.data;        n.next=n.next.next;        return true;        }            public static void main(String args[]) {        List list = new List(0);        list.appendToTail(1);        list.appendToTail(5);        list.appendToTail(3);        list.print();        deleteNode(list.next.next);        list.print();    }}

//结果{0, 1, 5, 3}{0, 1, 3}