单向链表的反转最简单方法

给出一个单向链表的header，要求经过处理变成反向，即原链表尾变为链表头，原链表头变成链表尾。

例如： 10->20->30->NULL

处理后变为： 30->20->10->NULL

我想，下面这应该是时间和空间方面都最简单的方法。

struct list{

int value;

struct list* next;

};

static int reverse(struct list **pl)

{

struct list* header,*tmp;

if(*pl==NULL) return 0;

header = NULL;

//add node to header and point to next node.

while(*pl!=NULL) {tmp=*pl; *pl=(*pl)->next; tmp->next=header; header=tmp;}

*pl = header;

return 1;

}

测试：

static void print_list(struct list *pl)

{

struct list *header = pl;

printf("list:");

while(header!=NULL){

printf(" %d",header->value);

header = header->next;

}

printf("/n");

}

int _tmain(int argc, _TCHAR* argv[])

{

#define NUMBER 9

int i = 0;

int score[NUMBER] = {10,20,30,40,50,60,70,80,90};

struct list *header,*cur,*tmp;

header = cur = NULL;

//create list.

while(i < NUMBER) {

tmp = (struct list *)malloc(sizeof(struct list));

tmp->value = score[i];

tmp->next = NULL;

if(i==0)

header = cur = tmp;

else{

cur->next = tmp;

cur = cur->next;

}

i++;

}

print_list(header);

reverse(&header);

print_list(header);

reverse(&header);

print_list(header);

printf("hello,world!/n");

getch();

return 0;

}

http://blog.csdn.net/knock/archive/2010/11/26/6036703.aspx