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算法之删除链表的重复的节点并返回头指针

作者:互联网

 

 分析和思路:使用map保存每个节点的个数,大于1的个数链表不创建,其他的重新创建,这个方法的缺点是用了o(n)的空间。

 1 /*
 2 struct ListNode {
 3     int val;
 4     struct ListNode *next;
 5     ListNode(int x) :
 6         val(x), next(NULL) {
 7     }
 8 };
 9 */
10 #include "iostream"
11 #include <map>
12 using namespace std;
13 //思路,先遍历整个链表,如果重复数量大于 1的都记录下来,然后再创建一个新的链表返回即可
14 
15 
16 
17 
18 
19 class Solution {
20 public:
21     ListNode* deleteDuplication(ListNode* pHead) {
22         map<int, int> m;
23         if (pHead == NULL)
24         {
25             return NULL;
26         }
27         ListNode* temp = NULL;
28         temp = pHead;
29         while (temp != NULL)
30         {
31             m[temp->val] += 1;
32             temp = temp->next;
33         }
34       
35         ListNode* pHeadBack = pHead;
36         temp = pHead;
37         while (temp!=NULL&&m[temp->val] > 1)
38         {
39             temp = temp->next;
40         }
41         if(temp==NULL)
42         {
43             return NULL;
44         }
45         pHeadBack = temp;
46         pHead = temp;
47         temp = temp->next;
48         ListNode* New;
49         while (temp != NULL)
50         {
51             if (m[temp->val] > 1)
52             {
53                 temp = temp->next;
54                 continue;
55             }
56             else
57             {
58                 New = (ListNode*)malloc(sizeof(ListNode));
59                 New->val = temp->val;
60                 New->next = NULL;
61                 pHead->next = New;
62                 New = New->next;
63                 
64                 pHead = pHead->next;
65                 temp = temp->next;
66             }
67         }
68         return pHeadBack;
69 
70     }
71 };

之前写的一个不用o(n)空间的,写的比较冗余,可读性比较差。

 1 /*
 2 struct ListNode {
 3     int val;
 4     struct ListNode *next;
 5     ListNode(int x) :
 6         val(x), next(NULL) {
 7     }
 8 };
 9 */
10 
11 class Solution {
12 public:
13     ListNode* deleteDuplication(ListNode* pHead)
14     {
15         if (pHead == NULL)
16         {
17             return NULL;
18         }
19         bool flag = false;
20         ListNode* q = pHead;
21         bool find_flag = false;
22         ListNode* back = (ListNode*)malloc(sizeof(ListNode));
23 
24         
25         if( pHead->next!=NULL)
26         {
27             if (pHead->val == pHead->next->val)
28             {
29                 back = pHead;
30 
31                 flag = true;
32 
33             }
34             
35         }
36         else
37         {
38             return pHead; //只有一个节点
39         }
40         
41         while(q->next!=NULL)
42         {
43             
44             if (find_flag == true)
45             {
46                 back->next = q->next;
47                 q = back;
48                 find_flag = false;
49             }
50             
51             if (q->val != q->next->val)
52             {
53                 back = q;
54 
55                 q = q->next;
56                 continue;
57             }
58             else
59             {
60                 while(q->val == q->next->val&&q->next!=NULL)
61                 { 
62                     q->next = q->next->next;
63                 }
64                 
65                 find_flag = true;
66                 //free(q) 释放节点
67                 continue;
68             }
69         }
70         if (flag == true)
71         {
72             if(find_flag ==true)
73                 {
74                    back->next = q->next;
75                     q = back;
76                     return pHead->next;
77                     
78                 }
79             else 
80                 return pHead->next;
81         }
82         else
83         {
84             return pHead;
85         }
86     
87     }
88 };

 

标签:ListNode,val,temp,next,链表,pHead,NULL,节点,指针
来源: https://www.cnblogs.com/technologykeeping/p/15915310.html