Cracking the coding interview--Q2.3

题目

原文：

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

EXAMPLE

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

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

译文：

实现一个算法去删除一个单链表中间的节点，只给出指向那个节点的指针

例如

输入：链表a->b->c->d->e中的节点c

结果：不用返回什么，只是新的链表变为a->b->d->e

解答

注意题目只给出要删除的节点，代码如下：

class Q2_3{

public static boolean deleteNode(Node n) {
if (n == null || n.next == null) {
return false; // Failure
}
Node next = n.next;
n.data = next.data;
n.next = next.next;
return true;
}

public static void main(String[] args){
int[] data=new int[]{1,2,3,2,5};

LinkList ll=new LinkList();
for(int i=0;i<data.length;i++){
ll.add(data[i]);
}

//ll.delete(3);
//ll.print();

deleteNode(ll.head.next.next);
ll.print();
}
}

class Node{
public Object data;
public Node next;

public Object getData() {
return data;
}
public void setData(Object data) {
this.data = data;
}
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
}

class LinkList{
public Node head;
public Node tail;
private int size;
public int getSize() {
return size;
}
public void setSize(int size) {
this.size = size;
}

public void add(Object i){//在链表结尾插入节点
Node newnode = new Node();
newnode.setData(i);
if(head == null){
head = newnode;
tail = newnode;
size ++ ;
}
else {
tail.setNext(newnode);
tail = newnode;
size ++ ;
}
}
//删除节点值为i的节点
public void delete(Object i){
if(head == null){
System.out.println("链表为空");
return;
}
Node temp = head;
while(temp.getNext().getData() != i){
temp= temp.getNext();
}
if(temp == null){
System.out.println("没有该节点");
return;
}
temp.setNext(temp.getNext().getNext());
size--;
}

public void print(){
if(head == null){
System.out.println("链表为空");
return;
}
Node temp = head;
while(temp.getNext()!= null){
System.out.println(temp.getData().toString());
temp=temp.getNext();
}
System.out.println(temp.getData().toString());
}
}


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