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This article will explain in detail how to use java to delete the repeated elements in the sorted list. The editor thinks it is very practical, so I share it with you as a reference. I hope you can get something after reading this article.

Given a sorted linked list, delete all nodes that contain duplicate numbers, leaving only those numbers that are not repeated in the original linked list.

Example 1:

Enter: 1-> 2-> 3-> 3-> 4-> 4-> 5

Output: 1-> 2-> 5

Example 2:

Enter: 1-> 1-> 1-> 2-> 3

Output: 2-> 3

Answer:

1public ListNode deleteDuplicates1 (ListNode head) {

2 if (head = = null) return null

3 ListNode FakeHead = new ListNode (0)

4 FakeHead.next = head

5 ListNode pre = FakeHead

6 ListNode cur = head

7 while (cur! = null) {

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

9 cur = cur.next

10}

11 if (pre.next = = cur) {

12 pre = pre.next

13} else {

14 pre.next = cur.next

15}

16 cur = cur.next

17}

18 return FakeHead.next

19}

Parsing:

Because it is sorting, this is easy. If the previous node is the same as the next node, make sure that the current node is deleted when you delete the next node. This problem is not very difficult, and there are many solutions. Let's take a look at a recursive solution.

1public ListNode deleteDuplicates2 (ListNode head) {

2 if (head = = null | | head.next = = null) return head

3 if (head.val! = head.next.val) {

4 head.next = deleteDuplicates2 (head.next)

5 return head

6} else {

7 while (head.next! = null & & head.val = = head.next.val)

8 head = head.next

9 return deleteDuplicates2 (head.next)

10}

11}

This kind of recursion is more concise, and the page is a little bit easier to understand.

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