【11_83】Remove Duplicates from Sorted List

这道题本质上不难，难的是细节处理，容易出错。

第一遍写的代码越改越大，越臃肿，此时，不如推倒重写，果然，第二次一遍过。

Remove Duplicates from Sorted List

My Submissions

Question

Total Accepted: 90731 Total Submissions: 255705 Difficulty: Easy

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

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

下面是Discuss里面的好代码：只有三行！！！

Java写法

1 public ListNode deleteDuplicates(ListNode head) {
2         if(head == null || head.next == null)return head;
3         head.next = deleteDuplicates(head.next);
4         return head.val == head.next.val ? head.next : head;
5 }

View Code

另一个好代码，和我的思想差不多，不过更简洁：

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

View Code

然后是我自己写的：

C语言

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 struct ListNode* deleteDuplicates(struct ListNode* head) {
 9     if (head == NULL)
10         return NULL;
11     if (head->next == NULL)
12         return head;
13
14     struct ListNode* p = head;
15     struct ListNode* q = p->next;
16     while(p->next != NULL)  {
17         if(p->val == p->next->val)  {
18             p->next = p->next->next;
19         }
20         else if(p->next != NULL)
21             p = p->next;
22     }
23
24     return head;
25 }

转载于:https://www.cnblogs.com/QingHuan/p/5051388.html

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