Leetcode: Remove Duplicates from Sorted List II 解题报告

Remove Duplicates from Sorted List II

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:

使用一个del标记来删除最后一个重复的字元。

 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) {

15             return null;

16         }

17         

18         // record the head.

19         ListNode dummy = new ListNode(0);

20         dummy.next = head;

21         

22         ListNode cur = dummy;

23         

24         // to delete the last node in the list of duplications.

25         boolean del = false;

26         

27         while (cur != null) {

28             if (cur.next != null

29                 && cur.next.next != null

30                 && cur.next.val == cur.next.next.val) {

31                 cur.next = cur.next.next;

32                 del = true;

33             } else {

34                 if (del) {

35                     cur.next = cur.next.next;

36                     del = false;

37                 } else {

38                     cur = cur.next;

39                 }

40             }

41         }

42         

43         return dummy.next;

44     }

45 }

View Code

SOLUTION 2:

使用一个pre, 一个cur来扫描，遇到重复的时候，使用for循环用cur跳过所有重复的元素。

 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) {

15             return null;

16         }

17         

18         ListNode dummy = new ListNode(0);

19         dummy.next = head;

20         

21         ListNode pre = dummy;

22         ListNode cur = pre.next;

23         

24         while (cur != null && cur.next != null) {

25             if (cur.val == cur.next.val) {

26                 while (cur != null && cur.val == pre.next.val) {

27                     cur = cur.next;

28                 }

29                 

30                 // delete all the duplication.

31                 pre.next = cur;

32             } else {

33                 cur = cur.next;

34                 pre = pre.next;

35             }

36         }

37         

38         return dummy.next;

39     }

40 }

View Code