笔试题&面试题：设计一个复杂度为n的算法找到单向链表倒数第m个元素

设计一个复杂度为n的算法找到单向链表倒数第m个元素.最后一个元素假定是倒数第0个.

提示:双指针查找

相对于双向链表来说,单向链表仅仅能从头到尾依次訪问链表的各个节点,所以假设要找链表的倒数第m个元素也仅仅能从头到尾进行查找,在查找的过程中,设定两个指针,当中p指针指向当前訪问的节点,q指针指向p之前的节点,且两者之间相距m个节点,这样,当p指针指向最后一个节点时,那q指针指向的元素就是倒数第m个元素,程序的处理步骤例如以下:

#include <stdio.h>
#include <malloc.h>

#define	NULL	0

typedef struct node
{
int data;
struct node *next;
}ElemSN;

void create_link(ElemSN *head, int num[], int len)
{
int i;

for(i = 0; i < len; i ++)
{
head->next = (ElemSN *)malloc(sizeof(ElemSN));
head->next->data = num[i];
head->next->next = NULL;
head = head->next;
}
}

void find_mlink(ElemSN *head, int m) /*最后一个单元为倒数第0个单元*/
{
ElemSN *p, *q;
int count = 0;

for(q = head, p = head->next; p; p = p->next)
{
count ++;
if(count > m)
{
q = q->next;
}
}

printf("The bottom number %d is: %d.\n", m, q->data);
}

void clear_link(ElemSN *head)
{
ElemSN *p, *q;

for(q = head, p = head->next; p; q = p, p = p->next)
{
free(q);
}
free(q);
}

int main()
{
ElemSN *head;
int num[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

head = (ElemSN *)malloc(sizeof(ElemSN));
head->data = NULL;
head->next = NULL;
create_link(head, num, 10);
find_mlink(head, 3); /*查找倒数第3个单元的值*/
clear_link(head);
head = NULL;

return 0;
}

程序执行截图：