LeetCode Collection (59) - 83 title Remove Duplicates from sorted list

problem

Given a sorted linked list, delete all duplicates such that each element appear only once. 

 Example 1: 


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


 Example 2: 


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

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translation:

Given a sorted list, remove all duplicate entries, so that each element occurs only once. Example 1: Input: 1 -> 1 -> Output 2: 1 -> 2 Example 2: Input: 1 -> 1 -> 2 -> 3 -> Output 3: 1 -> 2 -> 3

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Problem-solving ideas

This idea is very simple question, because the list is ordered, indicating if there are duplicate, certainly the next, traversing the order, the same as a node after the current node and if they are, then cover the current node. Traversal time to get

Problem-solving approach

1. According to our idea to edit the code as follows

   public ListNode deleteDuplicates(ListNode head) {
   
       if (head == null) {
           return head;
       }
   
       ListNode temp = head;
       while (head.next != null) {
           if (head.val == head.next.val) {
               head.next = head.next.next;
               continue;
           }
           head = head.next;
       }
       return temp;
   
   }
   
   public class ListNode {
       int val;
       ListNode next;
   
       ListNode(int x) {
           val = x;
       }
   }
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   Time Complexity : The program cycle with m so f (n) = (n) = n; so O (f (n)) = O (n), i.e., T (n) = O (n )

   Space complexity : The program uses the extra space is not used, so the space complexity is O (n) = O (1 );

to sum up

Solution of this problem is substantially above v, solving this problem is substantially only one way to directly read cover traversed, I by comparing the current node and the next node, and may be compared before a node.

Reproduced in: https: //juejin.im/post/5cfa8a846fb9a07ea803bc61