题解 | #输出单向链表中倒数第k个结点#
输出单向链表中倒数第k个结点
https://www.nowcoder.com/practice/54404a78aec1435a81150f15f899417d
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct Link {
int elem;
struct Link* next;
}link;
link* initlink(int n)
{
link* pnew, * head, * ptail;
pnew = (link*)malloc(sizeof(link));
scanf("%d", &pnew->elem);
head = pnew;
ptail = pnew;
for (int i = 1; i < n; i++)
{
pnew = (link*)malloc(sizeof(link));
scanf("%d", &pnew->elem);
ptail->next = pnew;
ptail = pnew;
}
ptail->next = NULL;
return head;
}
int main(void)
{
int n;
int k;
while (scanf("%d", &n) != EOF)
{
link* p;
link* p1;
p = initlink(n);
p1 = p;
scanf("%d", &k);
//计算链表长度
int cnt = 1;
while (p->next != NULL)
{
cnt++;
p = p->next;
}
int j;
j = cnt - k;
while (j != 0)
{
p1 = p1->next;
j--;
}
/*for (int j = 0; j < cnt; j++)
{
printf("%d", p1->elem);
p1 = p1->next;
}*/
printf("%d\n", p1->elem);
}
return 0;
}
