本文实例展示了C++实现输出链表中倒数第k个节点的方法，分享给大家供大家参考之用。
运行本文所述实例可实现输入一个单向链表，输出该链表中倒数第k个节点。
具体实现方法如下：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
|
/* * Copyright (c) 2011 alexingcool. All Rights Reserved. */#include <iostream>using namespace std;int array[] = {5, 7, 6, 9, 11, 10, 8};const int size = sizeof array / sizeof *array;struct Node{ Node(int i = 0, Node *n = NULL) : item(i), next(n) {} int item; Node *next;};Node* construct(int (&array)[size]){ Node dummy; Node *head = &dummy; for(int i = 0; i < size; i++) { Node *temp = new Node(array[i]); head->next = temp; head = temp; } return dummy.next;}void print(Node *head){ while(head) { cout << head->item << " "; head = head->next; }}Node* findKnode(Node *head, int k){ Node *pKnode = head; if(head == NULL) { cout << "link is null" << endl; return NULL; } while(k--) { if(head == NULL) { cout << "k is bigger than the length of the link" << endl; return NULL; } head = head->next; } while(head) { head = head->next; pKnode = pKnode->next; } return pKnode;}void main(){ Node *head = construct(array); cout << "source link: "; print(head); cout << endl; Node *kNode = findKnode(head, 5); if(kNode != NULL) cout << "the knode is: " << kNode->item << endl;} |
测试用例如下：
1. NULL Link
head = NULL;
2. normal Link, with normal k
k <= len(head);
3. normal Link, with invalid k
k > len(head)
希望本文所述对大家C程序算法设计的学习有所帮助。
