码迷,mamicode.com
首页 > 其他好文 > 详细

LeetCode82 Remove Duplicates from Sorted List II

时间:2016-10-08 23:32:28      阅读:195      评论:0      收藏:0      [点我收藏+]

标签:

题目:

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.(Medium)

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

分析:

链表去重的follow-up,但是题号这个在前面,要把有重复元素,则所有该元素均删掉。

注意:

1. 使用dummy node处理头结点被删除掉的情况。

2. 只要有head-> next -> val存在时,先判断head -> next是否为空。

代码:

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* deleteDuplicates(ListNode* head) {
12         ListNode dummy(0);
13         dummy.next = head;
14         head = &dummy;
15         while(head -> next != nullptr && head -> next -> next != nullptr) {
16             if (head -> next -> val == head -> next -> next -> val) {
17                 int val = head -> next -> val;
18                 while (head -> next != nullptr && head -> next -> val == val) { //只要取head->next,先判断
19                     ListNode* temp = head -> next;
20                     head -> next = head -> next -> next;
21                     delete temp;
22                 }
23             }
24             else {
25                 head = head -> next;
26             }
27         }
28         return dummy.next;
29     }
30 };

 

 

LeetCode82 Remove Duplicates from Sorted List II

标签:

原文地址:http://www.cnblogs.com/wangxiaobao/p/5940212.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!