6-4 Reverse Linked List(20 分)
阿新 • • 發佈:2018-12-24
6-4 Reverse Linked List(20 分)
Write a nonrecursive procedure to reverse a singly linked list in O(N) time using constant extra space.
Format of functions:
List Reverse( List L );
where List
is defined as the following:
typedef struct Node *PtrToNode; typedef PtrToNode List; typedef PtrToNode Position; struct Node { ElementType Element; Position Next; };
The function Reverse
is supposed to return the reverse linked list of L
, with a dummy header.
Sample program of judge:
#include <stdio.h> #include <stdlib.h> typedef int ElementType; typedef struct Node *PtrToNode; typedef PtrToNode List; typedef PtrToNode Position; struct Node { ElementType Element; Position Next; }; List Read(); /* details omitted */ void Print( List L ); /* details omitted */ List Reverse( List L ); int main() { List L1, L2; L1 = Read(); L2 = Reverse(L1); Print(L1); Print(L2); return 0; } /* Your function will be put here */
Sample Input:
5
1 3 4 5 2
Sample Output:
2 5 4 3 1
2 5 4 3 1
List Reverse(List L)
{if (L->Next == NULL || L->Next->Next==NULL)
return L;
else {
PtrToNode p = L->Next;
PtrToNode q = p->Next;
if (q->Next == NULL) {
p->Next = NULL;
q->Next = p;
}
else{
PtrToNode tmp = q->Next;
p->Next = NULL;
while (tmp) {
q->Next = p;
p = q;
q = tmp;
tmp = tmp->Next;
}
q->Next = p;
}
L->Next = q;
}
return L;
}