92. Reverse Linked List II

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

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

/**
 * 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 newhead = new ListNode(-1);
        newhead.next = head;
        
        if(head==null||head.next==null)
            return newhead.next;
            
        ListNode startpoint = newhead;//startpoint指向需要开始reverse的前一个
        ListNode node1 = null;//需要reverse到后面去的节点
        ListNode node2 = null;//需要reverse到前面去的节点
        
        for (int i = 0; i < n; i++) {
            if (i < m-1){
                startpoint = startpoint.next;//找真正的startpoint
            } else if (i == m-1) {//开始第一轮
                node1 = startpoint.next;
                node2 = node1.next;
            }else {
                node1.next = node2.next;//node1交换到node2的后面
                node2.next = startpoint.next;//node2交换到最开始
                startpoint.next = node2;//node2作为新的点
                node2 = node1.next;//node2回归到node1的下一个，继续遍历
            }
        }
        return newhead.next;
    }
    }

https://www.cnblogs.com/springfor/p/3864303.html

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) return null;
        ListNode dummy = new ListNode(-1);
        ListNode pre = dummy;
        dummy.next = head;
        for(int i = 0; i < m - 1; i++) {
            pre = pre.next;
        }
        
        ListNode cur = pre.next;
        for(int i = 0; i < n-m; i++) {
            ListNode move = cur.next;
            cur.next = move.next;
            move.next = pre.next;
            pre.next = move;            
        }
        return dummy.next;
    }
}

和25保持一致的写法，少记点幺蛾子屮