C语言实现输出链表中倒数第k个节点
投稿:shichen2014
这篇文章主要介绍了C语言实现输出链表中倒数第k个节点,主要涉及链表的遍历操作,是数据结构中链表的常见操作。需要的朋友可以参考下
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:
/* * 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。