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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。

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