链表反转的四种方法分析

关于反转链表的总结：
反转链表的方式有很多，现在介绍四种：
非递归的：









从图中可以看到，对n个节点的反转，方法一指针指向改变了2^(n-1)次，还有一个额外的节点，而方法二用了2^(n-1)次，却没有增加额外节点。方法三既没有增加新节点，指针指向改变了n次，显然方法三好。
递归方法很赞，图就不好画了，大家直接看代码。

代码：

package OfferTest;

import java.util.HashMap;

/**
* class ListNode
* @author cyc
*
*/
public class ListNode {
int val;
ListNode next;
ListNode(int x){
val = x;
}
/**
* print List from its head node
* @param head
*/
public static void printList(ListNode head){
HashMap<ListNode, Boolean> hm = new HashMap<ListNode, Boolean>();
ListNode mv = head;
while(mv!=null){
if(!hm.containsKey(mv)){
System.out.print(mv.val+"——>");
hm.put(mv, true);
mv = mv.next;
}
else{
System.out.print(mv.val+" is the circle List\n");
return;
}
}
System.out.print("null\n");
}
/**
* Transfer int[] array to List
* @param array
* @return ListNode
*/
public static ListNode arrayToList(int[] array){
if(array.length==0){
return null;
}
else{
ListNode head = new ListNode(array[0]);
ListNode end = head;
for(int i=1; i<array.length; i++){
ListNode tmpNode = new ListNode(array[i]);
end.next = tmpNode;
end = end.next;
}
end.next = null;
return head;
}
}
}

package OfferTest;

public class ReverseList {
public static void main(String[] args) {
int[] array = {1,2,3};
ListNode head = ListNode.arrayToList(array);
ReverseList rl = new ReverseList();
head = rl.reverseList1(head);
ListNode.printList(head);
head = rl.reverseList2(head);
ListNode.printList(head);
head = rl.reverseList3(head);
ListNode.printList(head);
head = rl.recusionList(head);
ListNode.printList(head);
}
/**
* 增加节点法
* @param head
* @return
*/
public ListNode reverseList1(ListNode head){
//增加一个新的节点newHead
ListNode newHead = new ListNode(0);
newHead.next = null;
ListNode cur;
while(head!=null){
cur = head;
head = head.next;
cur.next = newHead.next;
newHead.next = cur;
}
return newHead.next;
}
/**
* 反转两个节点的次序，需要指针移动两次
* @param head
* @return
*/
public ListNode reverseList2(ListNode head){
if(head==null||head.next==null){
return head;
}
ListNode cur = head.next;
ListNode end = head;
while(cur!=null){
end.next = cur.next;
cur.next = head;
head = cur;
cur = end.next;
}
return head;
}
/**
* 反转两个节点时，移动一次指针
* @param head
* @return
*/
public ListNode reverseList3(ListNode head){
if(head==null||head.next==null){
return head;
}
ListNode cur = head.next;
ListNode tmp;
head.next = null;
while(cur!=null){
tmp = cur.next;
cur.next = head;
head = cur;
cur = tmp;
}
return head;
}
/**
* 递归解法
* @param head
* @return
*/
public ListNode recusionList(ListNode head){
if(head==null||head.next==null){
return head;
}
ListNode nextHead = recusionList(head.next);
head.next.next = head;
head.next = null;
return nextHead;
}
}