The simplest way to reverse a one-way linked list

The header of a one-way linked list must be converted into a reverse link after processing, that is, the end of the original linked list is changed to the head of the linked list, and the head of the original linked list is changed to the end of the linked list.

Example: 10-> 20-> 30-> null

Changed to: 30-> 20-> 10-> null after processing

I think the following should be the simplest method in terms of time and space.

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;
}

Test:

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