LeetCode - 83, remove sorts the list of repeating elements

Given a sorted list, delete all the duplicate elements, so that each element occurs only once.

Example 1:

    Input: 1-> 1-> 2
    Output: 1-> 2

Example 2:

    Input: 1-> 1-> 2-> 3-> 3
    Output: 1-> 2-> 3

1  / * 
2  * list definition
 . 3  * {public class ListNode
 . 4  * int Val;
 . 5  * ListNode Next;
 . 6  * ListNode (int X) {X = Val;}
 . 7  *}
 . 8   * /

solution:

 1 class Solution {
 2     public ListNode deleteDuplicates(ListNode head) {
 3         //去掉特殊情况
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7         
 8         ListNode prev = head.next;
 9         ListNode end = head;
10         
11         while(prev != null) {
12             if (end.val == prev.val) {
13                 end.next = prev.next;
14                 prev = end.next;
15             } else {
16                 prev = prev.next;
17                 end = end.next;
18             }
19         }
20         
21         return head;
22     }
23 }

View Code

Description: Note input original list may be empty, to a special judge. The rest is simple list deletion.

Solution Optimization: just create a temporary variable

 1 public ListNode deleteDuplicates(ListNode head) {
 2     ListNode current = head;
 3     while (current != null && current.next != null) {
 4         if (current.next.val == current.val) {
 5             current.next = current.next.next;
 6         } else {
 7             current = current.next;
 8         }
 9     }
10     return head;
11 }

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Optimization Solution II: removing the wild pointer

 1 class Solution {
 2     public ListNode deleteDuplicates(ListNode head) {
 3         ListNode cur = head;
 4         while(cur != null && cur.next != null){
 5             if(cur.val == cur.next.val){
 6                 ListNode node = cur.next; 
 7                 cur.next = node.next;
 8                 node.next = null;//清除野指针
 9             }else{
10                 cur = cur.next;          
11             }
12             
13         }
14         return head;
15     }
16 }

View Code

Description (Personal understanding): ListNode = cur.next node; Next pointer still points to the node in the element cur.next.next garbage collection will not be recovered java