*92. Reverse Linked List II (follow up questions)-优快云博客

本文介绍了一种在链表中从位置m到n进行一次性、就地反转的方法，并提供了一个具体的示例来展示如何实现该算法。该算法适用于指定起点和终点的情况。

Reverse a linked list from position m to n. Do it in one-pass and in-place

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

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

FIrstly, I wanna use stack but need cosume space

Finally, I use in-plcae method

思路：reference: http://bangbingsyb.blogspot.com/2014/11/leetcode-reverse-linked-list-ii.html

反转整个链表的变种，指定了起点和终点。由于m=1时会变动头节点，所以加入一个dummy头节点

1. 找到原链表中第m-1个节点start：反转后的部分将接回改节点后。

从dummy开始移动m-1步

D->1->2->3->4->5->NULL

2. 将从p = start->next开始，长度为L = n-m+1的部分链表反转。

__________

| |

| V

D->1->2<-3<-4 5->NULL

| | |

st p h0

3. 最后接回

__________

| |

| V

D->1 2<-3<-4 5->NULL

|________|

In code

1. dfine dummy node

2. find the start node

3. reverse the node

4. connect the list

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode d = new ListNode(0);
        d.next = head;
        //find start point(head )
        ListNode pre = d; 
        for(int i = 0; i<m-1; i++){
            pre = pre.next;
            head = head.next;
        }
        ListNode tailReverse = head;
        ListNode end = null, temp;
        for(int i = 0; i<n-m+1; i++){ // need +1
            temp = head.next;
            head.next = end;
            end = head;
            head = temp;
        }
        //System.out.println(end.val);
        pre.next = end;
        tailReverse.next = head;
        return d.next; 
    }
}