ReverseLinkedList(翻转链表)

链表是一种物理存储单元上非连续、非顺序的存储结构,数据元素的逻辑顺序是通过链表中的指针链接次序实现的。非连续、非顺序指的是,通过指针把一组零散的内存块串联在一起,其中每一个内存块叫做链表的节点,所以每个节点包含两部分一个data(你存放的数据),一个next(指向下一个节点)当然这里使用单向链表举例。双向链表则有两个指向节点一个next(下一个节点)一个prev(上一个节点),对比:双向链表虽然更麻烦,但是比单向链表更受欢迎,因为,他记录了上一个节点,节省了操作数据时候的时间,但是需要用内存换时间。

 单向链表:

双向链表;

题目一:

给定一组数据,从尾到头进行翻转

Input:  1 ->2->3->4->5

Output: 5->5->3->2->1

思路:使用是三个变量进行控制(prev、current、next),模拟一个窗口,进行对数据的推进,进而翻转,不断改变三个元素的位置。

/**
* @author : lizi
* @date : 2021-03-02 17:05
**/
public class ListNode {
int val = 0;
ListNode next; ListNode(int x) {
val = x;
} public void setVal(int val) {
this.val = val;
} public void setNext(ListNode next) {
this.next = next;
}
}
    private static ListNode reverseList(ListNode head) {
if (head == null) return null;
ListNode prev=head;
ListNode current=head.next;
// because the last Node point into null,so let fist node.next point into null
prev.next=null;
while (current!=null){
ListNode next=current.next;
// that is place where change node. change position of current and prev
current.next=prev;
prev=current;
// change current into next (in the way,the window is pushed)
current=next;
}
return prev;
}
}

题目二:

obviously,we should change position of m and n 

Input:1->2->3->4->5->null ,m=2,n=5

Output:1->4>3>2>5>null 

tip: actually, many companies like to make this question as ultimate question,also the question was Amazon’s question for  interviewee,however, we remain can use method in reversing whole linkedList. i mean that method of pushing window.

    private static ListNode reverseBetween(ListNode head, int m, int n) {
// if you can not find head, you can make the node.next to find head, todo actually, it is a node that prevent you can not find first node
ListNode dummy = new ListNode(-1);
dummy.next=head;
head=dummy;
// to find a place where you began to control list
for (int i = 1; i <m ; i++) {
head=head.next;
}
ListNode prevM=head;
ListNode mNode=prevM.next;
ListNode nNode=mNode;
ListNode postN=nNode.next;
//todo visibly, the portion is method of window that i mention in method of reversing the whole window
for (int i = m; i <n ; i++) {
ListNode next=postN.next;
postN.next=nNode;
nNode=postN;
postN=next;
}
prevM.next=nNode;
mNode.next=postN;
// you can get dummy node.next in this way ,find the first node
return dummy.next;
}

finally: to do examine

    public static void main(String[] args) {
ListNode listNode1 = generateDate();
System.out.println(listNode1.toString());
// reverse whole list
ListNode reverseBetween = reverseList(generateDate());
// reverse m and n in list
ListNode listNode = reverseBetween(listNode1, 2, 4);
printForSomething(reverseBetween);
} // just to print result
public static void printForSomething(ListNode listNode) {
if (listNode != null) {
System.out.println(listNode.val);
printForSomething(listNode.next);
}
}

the picture is what i visualize method how to run:

To summarize:

as a beginner,personally, it is abstract to imagine how the cell of linkedlist  connect with each other,but it is a effective to draw what you imagine. i still want everyone to criticize what i wrong in every article.

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