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LeetCode: Remove Duplicates from Sorted List

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.

地址:https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/

算法:不算什么难题,直接看代码:

 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         if(!head || head->next == NULL) return head;
13         ListNode *p = head;
14         while(p->next){
15             if(p->next->val == p->val){
16                 ListNode *q = p->next;
17                 p->next = q->next;
18                 free(q);
19             }else{
20                 p = p->next;
21             }
22         }
23         return head;
24     }
25 };

第二题:

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.

地址:https://oj.leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

算法:这一题比上面一题稍微麻烦一点,但也不算很难,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         if(!head || head->next == NULL) return head;
13         ListNode *pre = NULL;
14         ListNode *p = head;
15         while(p){
16             ListNode *q = p->next;
17             while(q && q->val == p->val){
18                 q = q->next;
19             }
20             if(p->next == q){
21                 pre = p;
22                 p = p->next;
23             }else{
24                 if(pre){
25                     pre->next = q;
26                 }else{
27                     head = q;
28                 }
29                 while(p != q){
30                     ListNode *tempP = p->next;
31                     free(p);
32                     p = tempP;
33                 }
34                 p = q;
35             }
36         }
37         return head;
38     }
39 };

 

posted on 2014-09-03 21:57  Boostable  阅读(145)  评论(0编辑  收藏  举报