leetcode 203:Remove Linked List Elements
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题目:
Remove all elements from a linked list of integers that have value val.
Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5
分析:
本题考查链表的操作,要求删除链表元素,只要将指向被删除元素的指针指向被删除元素的后面一个元素就行,在java中要实现这种操作,必须先弄明白对象和引用的概念和关系,java中的引用类似于指针的概念。
代码:
public class LinkedList { /** * 链表结点类 * @author Don * */ public static class ListNode{ int val; ListNode next; ListNode(int x){val=x;} } /** * 删除链表元素 * @param head * @param val * @return */ public static ListNode removeElements(ListNode head,int val){ ListNode dummy=new ListNode(0); dummy.next=head; ListNode p=dummy; ListNode q=head; while(q!=null){ if(q.val==val){ p.next=q.next; }else{ p=p.next; } q=q.next; } return dummy.next; } /** * 构建单链表 * @param node * @param val */ public static ListNode buildLinkedList(int[] a){ ListNode before=null; for(int i=0;i<a.length;i++){ ListNode node=new ListNode(a[i]); node.next=before; before=node; } return before; } /** * 打印单链表 * @param node */ public static void show(ListNode head){ while(head!=null){ System.out.print(head.val+"."); head=head.next; } } public static void main(String[] args){ int[] a={1,2,3,4,5}; ListNode head=buildLinkedList(a); show(head); System.out.println(); ListNode temp=removeElements(head,5); show(temp); }}
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