【LeetCode】C# 92、Reverse Linked List II

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

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

给定链表和两个数值，将链表中对应两个位置的那段链表反转。

思路：建立4个指针。dummy指向head，pre指向第一个要翻转的节点，然后start指向下一个操作要翻转到前面的节点。then指向start.next，用来辅助翻转。

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution {    public ListNode ReverseBetween(ListNode head, int m, int n) {        if(head == null) return null;        ListNode dummy = new ListNode(0);         dummy.next = head;        ListNode pre = dummy;        for(int i = 0; i<m-1; i++) pre = pre.next;        ListNode start = pre.next;         ListNode then = start.next;         for(int i=0; i<n-m; i++)        {            start.next = then.next;            then.next = pre.next;            pre.next = then;            then = start.next;        }        return dummy.next;    }}