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;

}