/* reverse 1 -> 2 -> 3 -> 4 to 4 -> 3 -> 2 -> 1
* notice there is no head node
*
* reverse while traversal
*
* The reverse method traversals the linked list once,
* the time complexity is O(N).
*
* This solution preserves original linked list
* the space complexity is O(N).
*
*
* */
public class ExtendSolutionTwo {
public static MyNode reverse(MyNode head) {
if(head == null || head.next == null) return head;
//special treatment for tail
MyNode pre = new MyNode(head.value, null);
MyNode current = pre;
while(head.next != null) {
//create a new node, pointing to previous node
current = new MyNode(head.next.value, current);
head = head.next;
}
return current;
}
public static void main(String...args) {
MyNode head = createLinkedList();
MyNode preHead = new MyNode(head.value, head.next);
printResult(head);
head = reverse(head);
printResult(head);
printResult(preHead);
}
public static void printResult(MyNode head) {
System.out.println();
if(head == null) return;
while(head != null) {
System.out.print(head.value + " ");
head = head.next;
}
}
public static MyNode createLinkedList() {
MyNode head = new MyNode(1, null);
MyNode current = head;
for(int i = 2; i <= 4; i++) {
current.next = new MyNode(i, null);
current = current.next;
}
return head;
}
}
