每天一个小算法（4）----在O(1)时间删除指定结点

O(1)时间内删除结点的思路只能是复制该结点下一个结点的数据，然后删除该结点的下一个结点，来等效删除此结点。

需要注意的地方是删除头结点和尾结点的处理。

 1 #include <stdio.h>
 2 #include <time.h>
 3 #include <stdlib.h>
 4 typedef struct Node
 5 {
 6     int data;
 7     Node* next;
 8 }Node, *List;
 9 
10 List createList(int num) //随机生成数字，构造链表
11 {
12     List aList = (List)malloc(sizeof(Node));
13     aList->next = NULL;
14     aList->data = 0;
15     Node* qT = aList;
16 
17      // srand((int)time(0));
18      for ( int i=0; i< num; ++i)
19      {
20          Node* pTN = (Node*)malloc(sizeof(Node));
21          pTN->data = rand()%100;
22          pTN->next = NULL;
23          qT->next = pTN;
24          qT = pTN;
25      }
26      return aList;
27 }
28 
29 void printList(List aList)    //打印链表
30 {
31     if ( aList == NULL || aList->next == NULL )
32         return;
33 
34     Node* pT = aList->next;
35     printf("element of list:\n\t");
36     while( pT != NULL )
37     {
38         printf("%d ", pT->data);
39         pT = pT->next;
40     }
41 
42     printf("\n");
43 }
44 
45 void deleteList(List aList)    //删除链表
46 {}
47 
48 
49 //删除结点主算法
50 void deleteNode(List aList, Node* pNode)
51 {
52     if ( aList == NULL || pNode == NULL )
53         return;
54 
55     if ( aList == pNode )
56     {
57         printf("refuse to delete head node\n");
58         return;
59     }
60 
61     if ( pNode->next == NULL )
62     {
63         Node* pT = aList->next;
64         while ( pT->next != NULL )
65         {
66             pT = pT->next;
67         }
68 
69         delete pNode;
70         pT->next == NULL;
71     }
72 
73     Node* pN = pNode->next;
74     pNode->data = pNode->next->data;
75     pNode->next = pNode->next->next;
76 
77     delete pN;
78 }
79 
80 int main(int argc, char const *argv[])
81 {
82      srand((int)time(0));
83     List aList = createList(5);
84     printList(aList);
85     deleteNode(aList,aList->next->next);
86     printList(aList);
87     deleteList(aList);
88     
89     return 0;
90 }

 

posted @ 2014-06-13 14:43  Yrpen  阅读(433)  评论(0)  编辑  收藏  举报