问题10：还是单链表的一些问题（如单链表，怎样只遍历一次就可以求出中间结点）

问题描述：

给出一个单链表，不知道结点N的值，怎样只遍历一次就可以求出中间结点？

/*
*该题的大致思路是设置两个指针，比如*p1,*p2。p2每次移动两个位置，而p1每次移动一个位置。当p2移动到最后时，p1就是中间结点了。
*/
node *searchMid(node *head)
{
if(NULL == head)
return head;
if(NULL == head->next)
return head->next;
node *p1, *p2;
P1 = P2 = head;
while(p2->next!= NULL)
{
/*p2结点移动2个结点位置*/
P2 = P2->next;
if(p2->next != NULL)
p2 = p2->next;
p1 = p1->next;
}
return p1;
}

单链表插入结点

       关于单链表插入结点，大体要注意三种情况：

一是在第一个结点前插入；

二是在链表的中间插入；

三是在链表的最后插入。

另外还需要注意单链表没有指向前驱结点的指针，所以在实际程序中往往需要记录前驱结点，相信只要知道了这几点写出相应的代码就不太困难了。

代码如下：

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

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

//创建链表
node *creat()
{
node *head, *p, *s;
int x, cycle = 1;
head = (node *)malloc(sizeof(node));
p = head;
while(cycle)
{
printf("\nplease enter the data:");
scanf("%d", &x);
if(x != 0)
{
s = (node *)malloc(sizeof(node));
s->data = x;
printf("\n%d", s->data);
p->next = s;
p = s;
}
else
cycle = 0;
}
head = head->next;
printf("\nThe value of the first node is:%d\n", head->data);
p->next = NULL;
return head;
}

//单链表插入结点
node *insert_to_list(node *head, int num)
{
node *cur, *pre, *s;
cur = head;
s = (node *)malloc(sizeof(node));
s->data = num;
while(s->data > cur->data && cur->next != NULL)
{
pre = cur;
cur = cur->next;
}
if(s->data <= cur->data)
{
if(cur == head)
{
s->next = head;
head = s;
}
else
{
s->next = cur;
pre->next = s;
}
}
else
{
cur->next = s;
s->next = NULL;
}

return head;
}

//单链表测长
int length(node *head)
{
int len = 0;
node *p = head;
while(p != NULL)
{
len++;
p = p->next;
}
return len;
}

//单链表打印
void print(node *head)
{
node *p = head;
int n = length(head);
printf("\nNow, These %d records are:\n", n);
while(p != NULL)
{
printf("\n uuu %d ", p->data);
p = p->next;
}
}

int main(void)
{
//creat a list
node *List = creat();
print(List);
node *p = insert_to_list(List, 5);
print(p);
return 0;
}

运行结果：



编程实现单链表的逆置

代码如下：

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

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

//创建链表
node *creat()
{
node *head, *p, *s;
int x, cycle = 1;
head = (node *)malloc(sizeof(node));
p = head;
while(cycle)
{
printf("\nplease enter the data:");
scanf("%d", &x);
if(x != 0)
{
s = (node *)malloc(sizeof(node));
s->data = x;
printf("\n%d", s->data);
p->next = s;
p = s;
}
else
cycle = 0;
}
head = head->next;
printf("\nThe value of the first node is:%d\n", head->data);
p->next = NULL;
return head;
}

//单链表插入结点
node *insert_to_list(node *head, int num)
{
node *cur, *pre, *s;
cur = head;
s = (node *)malloc(sizeof(node));
s->data = num;
while(s->data > cur->data && cur->next != NULL)
{
pre = cur;
cur = cur->next;
}
if(s->data <= cur->data)
{
if(cur == head)
{
s->next = head;
head = s;
}
else
{
s->next = cur;
pre->next = s;
}
}
else
{
cur->next = s;
s->next = NULL;
}

return head;
}

//单链表测长
int length(node *head)
{
int len = 0;
node *p = head;
while(p != NULL)
{
len++;
p = p->next;
}
return len;
}

//单链表打印
void print(node *head)
{
node *p = head;
int n = length(head);
printf("\nNow, These %d records are:\n", n);
while(p != NULL)
{
printf("\n uuu %d ", p->data);
p = p->next;
}
}

//实现单链表的逆置
node *reverse(node *head)
{
node *p1, *p2, *p3;
if(head == NULL || head->next == NULL)
return head;
p1 = head;
p2 = head->next;
while(p2)
{
p3 = p2->next;
p2->next = p1;
p1 = p2;
p2 = p3;
}
head->next = NULL;
head = p1;
return head;
}

int main(void)
{
//creat a list
node *List = creat();
print(List);
node *p = insert_to_list(List, 5);
print(p);
p = reverse(p);
print(p);
return 0;
}

运行结果：