C语言实现输出链表中倒数第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; } 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;}