每天一个小算法（5）----找到链表倒数第K个结点

估计这个问题在面试中被问烂了。

思路是先找到正数的第K个结点的指针pT，然后和指向头结点的指针pN一起向后移动，直到第K个指针指向NULL，此时pN指向的结点即倒数第K个结点。

如图：

 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 
11 List createList(int num) //随机生成数字，构造链表
12 {
13     List aList = (List)malloc(sizeof(Node));
14     aList->next = NULL;
15     aList->data = 0;
16     Node* qT = aList;
17 
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)    //正序打印链表
30 {
31     if ( aList == NULL || aList->next == NULL )
32         return;
33 
34     Node* pT = aList->next;
35     printf("element of the 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 //链表倒数第K个节点主算法
46 Node* rfind_K_node(List aList, int K)
47 {
48     if ( aList == NULL || K <=0 )
49         return NULL;
50 
51     Node* pT = aList;
52     Node* pN = aList;
53     for ( int i=0; i< K; ++i )
54     {
55         pT = pT->next;
56     }
57 
58     while ( pT != NULL )
59     {
60         pN = pN->next;
61         pT = pT->next;
62     }
63 
64     return pN;
65 }
66 
67 void deleteList(List aList)    //删除链表
68 {}
69 
70 int main(int argc, char const *argv[])
71 {
72      srand((int)time(0));
73     List aList = createList(7);
74     printList(aList);
75     printf("The bottom %d Node data is %d\n", 5, rfind_K_node(aList, 5)->data);
76     deleteList(aList);
77     
78     return 0;
79 }

 

 

posted @ 2014-06-14 15:42  Yrpen  阅读(747)  评论(1)  编辑  收藏  举报