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.
Solution:
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) return head; 15 ListNode preHead = new ListNode(-1); 16 preHead.next = head; 17 ListNode cur = head; 18 ListNode pre = preHead; 19 20 //NOTE:We need to be every carefull about the condition, 21 //because if the last node be deleted, the cur is null in the next step. 22 //however, if the node before the last node is deleted, the cur is the last node in the next step, 23 //we then need to stop also, otherwise cur.next.val is invalid. 24 while (cur!=null && cur.next!=null){ 25 if (cur.val==cur.next.val){ 26 while (cur.val==cur.next.val){ 27 cur.next = cur.next.next; 28 if (cur.next==null) 29 break; 30 } 31 pre.next = cur.next; 32 cur = pre.next; 33 } else { 34 pre = cur; 35 cur = cur.next; 36 if (cur.next==null) 37 break; 38 } 39 } 40 41 return preHead.next; 42 } 43 }