Cracking the coding interview--Q2.2

题目

原文：

Implement an algorithm to find the nth to last element of a singly linked list.

译文：

实现一个算法找出一个单链表的倒数第n个元素。

解答

代码如下：

import java.util.*;

class Q2_2{
public static Object nthtoLastNode(LinkList ll,int n) {
Node head=ll.head;
Node node = head;
int count=n-1;
while (count >= 1 && node.next != null) {
count--;
node = node.next;
}
//链表的长度小于n
if (count != 0) return null;

while (node.next != null) {
head = head.next;
node = node.next;
}
return head.data;
}
public static void main(String[] args){
int[] data=new int[]{1,2,3,2,5,2};

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

System.out.println(nthtoLastNode(ll,4));
}
}

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 ++ ;
}
}
}

---EOF---