java怎么在双向链表中删除元素？

假如我创建一个双向链表 a
存储的data是String
“a”-"b"-"c"
这个样子，a的size是3
我写了一个remove的程序， 让 b.previoud.next = b.next;
如果输出新的链表a， 结果是 “a”-“c”
但是a的size仍然是3，请问怎么办？

你自己写的remove方法吗？
在你写的remove方法里size--

双向链表：就是有双向指针，即双向的链域。 链结点的结构： ┌────┬────┬────────┐ │ data │ next │ previous │ └────┴────┴────────┘ 双向链表不必是双端链表(持有对最后一个链结点的引用)，双端链表插入时是双向的。 有两条链：一条从头到尾，一条从尾到头，删除遍历时也是双向的。 public class doublylinkedlist { private link head; //首结点 private link rear; //尾部指针 public doublylinkedlist() { } public t peekhead() { if (head != null) { return head.data; } return null; } public boolean isempty() { return head == null; } public void insertfirst(t data) {// 插入 到 链头 link newlink = new link(data); if (isempty()) {//为空时，第1次插入的新结点为尾结点 rear = newlink; } else { head.previous = newlink; //旧头结点的上结点等于新结点 } newlink.next = head; //新结点的下结点旧头结点 head = newlink; //赋值后，头结点的下结点是旧头结点，上结点null } public void insertlast(t data) {//在链尾 插入 link newlink = new link(data); if (isempty()) { head = newlink; } else { rear.next = newlink; } newlink.previous = rear; rear = newlink; //赋值后，尾结点的上结点是旧尾结点，下结点null } public t deletehead() {//删除 链头 if (isempty()) return null; link temp = head; head = head.next; //变更首结点，为下一结点 if (head != null) { head.previous = null; } else { rear = null; } return temp.data; } public t deleterear() {//删除 链尾 if (isempty()) return null; link temp = rear; rear = rear.previous; //变更尾结点，为上一结点 if (rear != null) { rear.next = null; } else { head = null; } return temp.data; } public t find(t t) {//从头到尾find if (isempty()) { return null; } link find = head; while (find != null) { if (!find.data.equals(t)) { find = find.next; } else { break; } } if (find == null) { return null; } return find.data; } public t delete(t t) { if (isempty()) { return null; } link current = head; while (!current.data.equals(t)) { current = current.next; if (current == null) { return null; } } if (current == head) { head = head.next; if (head != null) { head.previous = null; } } else if (current == rear) { rear = rear.previous; if (rear != null) { rear.next = null; } } else { //中间的非两端的结点，要移除current current.next.previous = current.previous; current.previous.next = current.next; } return current.data; } public boolean insertafter(t key, t data) {//插入在key之后, key不存在return false if (isempty()) { return false; } link current = head; while (!current.data.equals(key)) { current = current.next; if (current == null) { return false; } } link newlink = new link(data); if (current == rear) { rear = newlink; } else { newlink.next = current.next; current.next.previous = newlink; } current.next = newlink; newlink.previous = current; return true; } public void displaylist4head() {//从头开始遍历 system.out.println("list (first-->last):"); link current = head; while (current != null) { current.displaylink(); current = current.next; } } public void displaylist4rear() {//从尾开始遍历 system.out.println("list (last-->first):"); link current = rear; while (current != null) { current.displaylink(); current = current.previous; } } class link {//链结点 t data; //数据域 link next; //后继指针，结点 链域 link previous; //前驱指针，结点 链域 link(t data) { this.data = data; } void displaylink() { system.out.println("the data is " + data.tostring()); } } public static void main(string[] args) { doublylinkedlist list = new doublylinkedlist(); list.insertlast(1); list.insertfirst(2); list.insertlast(3); list.insertfirst(4); list.insertlast(5); list.displaylist4head(); integer deletehead = list.deletehead(); system.out.println("deletehead:" + deletehead); list.displaylist4head(); integer deleterear = list.deleterear(); system.out.println("deleterear:" + deleterear); list.displaylist4rear(); system.out.println("find:" + list.find(6)); system.out.println("find:" + list.find(3)); system.out.println("delete find:" + list.delete(6)); system.out.println("delete find:" + list.delete(1)); list.displaylist4head(); system.out.println("----在指定key后插入----"); list.insertafter(2, 8); list.insertafter(2, 9); list.insertafter(9, 10); list.displaylist4head(); } }