Leetcode: 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.

难度：70，参考，现在要把前驱指针指向上一个不重复的元素中，如果找到不重复元素，则把前驱指针知道该元素，否则删除此元素。算法只需要一遍扫描，时间复杂度是O(n)，空间只需要几个辅助指针，是O(1)

程序中prev指的元素定义为是上一个不重复的元素，最开始指向dummy node, cur指向定义为下一个元素为我们需要的不重复元素，最开始指向head。最后删除重复元素就是删除(prev，cur]区间内的元素，只需要prev.next = cur.next。而判断前后两个元素是否是重复只需要每次比较prev.next与cur.next, 如果重复，那么就把cur=cur.next一直往后移直到prev.next != cur.next或者到最末。如果发生了重复删除，prev不动，cur往后移动一位，否则没有重复prev和cur都要往后移一位，判断是否发生了删除就是看prev.next是否就是cur(prev.next == cur? 注意这里是内存地址一致，不是仅仅值相等)

比较好的做法：

 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         ListNode dummy = new ListNode(-1);

15         dummy.next = head;

16         ListNode cur = head;

17         ListNode prev = dummy;

18         while (cur != null) {

19             while (cur.next != null && prev.next.val == cur.next.val) { //compare prev.next and cur.next to see if duplicates

20                 cur = cur.next; //if duplicates, cur keeps moving one step furthur until cur.next != prev.next, (prev, cur] should be deleted 

21             }

22             if (prev.next == cur) { //to see if there has been duplicate or not

23                 prev = prev.next;  //no duplicates, prev and cur should move 1 step furthur together

24             }

25             else {

26                 prev.next = cur.next;  //duplicates exists, delete the duplicates, only moves cur 1 step further, prev stays.

27             }

28             cur = cur.next;

29         }

30         return dummy.next;

31     }

32 } 

另一种做法：第14行必须要在runner.next!=null的情况下才敢runner= runner.next, 否则runner就会指向null，下一次while循环判断runner.next!=null就会出错

 1 public class Solution {

 2     public ListNode deleteDuplicates(ListNode head) {

 3         if (head == null) return null;

 4         ListNode dummy = new ListNode(-1);

 5         dummy.next = head;

 6         ListNode walker = dummy;

 7         ListNode runner = head;

 8         while (runner.next != null) {

 9             if (walker.next.val == runner.next.val) {

10                 while (runner.next!=null && walker.next.val == runner.next.val) {

11                     runner = runner.next;

12                 }

13                 walker.next = runner.next;

14                 if (runner.next != null) runner = runner.next;

15             }

16             else {

17                 walker = walker.next;

18                 runner = runner.next;

19             }

20         }

21         return dummy.next;

22     }

23 }