求单链表倒数第k个结点

题目：

        输入一个单向链表，输出该链表中倒数第k个结点，链表的倒数第0个结点为链表的尾指针。

分析：

      设置两个指针p1，p2。首先p1和p2都指向head。然后p2向前走k步，这样p1和p2之间就间隔k个节点，然后p1和p2同....

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;


struct ListNode{
char data;
ListNode* next;
};

ListNode* head, *p, *q;
ListNode* pone, *ptwo;
ListNode* fun(ListNode* head, int k)
{
pone = ptwo = head;
for (int i = 0; i <= k - 1; i++)
ptwo = ptwo->next;
while (ptwo != NULL)
{
pone = pone->next;
ptwo = ptwo->next;
}
return pone;
}
int main()
{
char c;
c = getchar();
head = (ListNode*)malloc(sizeof(ListNode));
head->next = NULL;
p = head;
while (c != '0')
{
q = (ListNode*)malloc(sizeof(ListNode));
q->data = c;
q->next = NULL;
p->next = q;
p = p->next;
c = getchar();
}
cout << "--------" << endl;
cout << fun(head, 2)->data << endl;
return 0;
}