C语言实现获取链表中倒数第k个节点

/* 
* 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;
 }

 // 快指针先走 k 步
 while(k--) {
 if(head == NULL) {
  cout << "k is bigger than the length of the link" << endl;
  return NULL;
 }
  head = head->next;
 }

 // 双指针同步移动
 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;
}