Given theheadof a singly linked list and two integersleftandrightwhereleft <= right, reverse the nodes of the list from positionleftto positionright, and returnthe reversed list.

Example 1:

Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]

Example 2:

Input: head = [5], left = 1, right = 1
Output: [5]

Constraints:

• The number of nodes in the list isn.
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 
 
*/
class Solution {
    public ListNode reverseBetween(ListNode head, int l, int r) {
        ListNode pre = new ListNode(0);
        pre.next = head;
        ListNode leftpre = pre,rightpre=pre;
        //leftpre指向left的前一個
        for(int i=1;i<l;i++)  leftpre=leftpre.next;
        //rightpre指向right當前，記住這裡是當前，不是前一個！
 

        for(int i=1;i<=r;i++) rightpre=rightpre.next;
        //left指向要反轉的第一個node
        ListNode left=leftpre.next;
        //right指向第一個不反轉的node
        ListNode right = rightpre.next;
        //防止死迴圈
        rightpre.next=null;
        //反轉前的第一個為反轉後的最後一個
        ListNode lefttail = left;
        //進行反轉，並將反轉結果對接leftpre
        leftpre.next = reverse(left);
        //對接反轉自後一個與未反轉部分
        lefttail.next=right;
        return pre.next;
    }
    //簡單的連結串列反轉過程
    public ListNode reverse(ListNode head){
        ListNode pre = new ListNode(0);
        while(head!=null){
            ListNode temp=pre.next;
            pre.next=head;
            head=head.next;
            pre.next.next=temp;
        }
        return pre.next;
    }
}