Cracking coding interview(2.2)返回单向链表的倒数第n个节点

来源：互联网发布：网络弱电施工标准编辑：程序博客网时间：2024/06/08 07:20

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

1.先计算单向链表长度，再查找倒数第n个元素

2.使用2个Node节点指针pre，p，使pre指向head，p指向第n个节点，之后同步向后同步移动到p到达单向链表的最后一个节点为止，此时pre的值即指向单向链表倒数第n节点。(此方法一开始没想到，很妙！)

class Node{int data;Node next;public Node(int data, Node next){this.data = data;this.next = next;}}public class Solution{//time complexity:O(n), generally thinkingpublic static Node getLastN1(Node head, int n){if(n <= 0)return null;//counter length of singly linked listint len=0;Node p = head;for(;p != null;p = p.next)len++;if(len < n)return null;elselen -= n;//find last n nodep = head;for(;p != null;p = p.next)if(len == 0)return p;elselen--;return null;}//time complexity:O(n), optimum solutionpublic static Node getLastN2(Node head, int n){if(n <= 0)return null;Node p = head;//step forward n-1 stepsfor(;n != 1 && p != null;p = p.next){n--;}if(p == null)return null;//pre and p synchronization forwardNode pre = head;for(;p.next != null;p = p.next, pre = pre.next);return pre;}private static void printNode(Node head){Node p = head;for(;p != null;p = p.next)System.out.print(p.data+" ");System.out.println();}public static void main(String[] args){System.out.println("args.length="+args.length);Node head = new Node(1, null);head.next = new Node(2, null);head.next.next = new Node(3, null);head.next.next.next = new Node(4, null);head.next.next.next.next = new Node(5, null);head.next.next.next.next.next = new Node(6, null);head.next.next.next.next.next.next = new Node(7, null);Solution.printNode(head);Node p = Solution.getLastN1(head, 0);System.out.println("n=0:"+ (p == null ? p : p.data));p = Solution.getLastN1(head, 3);System.out.println("n=3:"+ (p == null ? p : p.data));p = Solution.getLastN1(head, 1);System.out.println("n=1:"+ (p == null ? p : p.data));p = Solution.getLastN1(head, 17);System.out.println("n=17:"+ (p == null ? p : p.data));Solution.printNode(head);p = Solution.getLastN2(head, 0);System.out.println("n=0:"+ (p == null ? p : p.data));p = Solution.getLastN2(head, 3);System.out.println("n=3:"+ (p == null ? p : p.data));p = Solution.getLastN2(head, 1);System.out.println("n=1:"+ (p == null ? p : p.data));p = Solution.getLastN2(head, 17);System.out.println("n=17:"+ (p == null ? p : p.data));}}

0 0