Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
每个节点与前驱节点还有后继相比较，若有一个相等，就跳过这个节点。
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public ListNode deleteDuplicates(ListNode head) { 14 if (head == null || head.next == null) { 15 return head; 16 } 17 ListNode start = new ListNode(head.val - 1); 18 ListNode temp=start; 19 ListNode curr=head; 20 int prev=start.val; 21 int next=head.next.val; 22 while (curr.next != null) { 23 if (curr.val != prev && curr.val != curr.next.val) { 24 prev = curr.val; 25 temp.next = curr; 26 temp=temp.next; 27 curr = curr.next; 28 } else { 29 prev=curr.val; 30 curr=curr.next; 31 if(curr.next!=null) next=curr.next.val; 32 } 33 } 34 if (curr.val != prev) { 35 temp.next = curr; 36 }else temp.next=null; 37 return start.next; 38 39 } 40 }