【Leetcode刷题篇】- Leetcode092反转链表II_ds单链表反转链表,给定一个链表,以及输入两个整数left和right

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例 1：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
示例 2：

输入：head = [5], left = 1, right = 1
输出：[5]

反转链表记得用伪结点

/**
 * 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 left, int right) {
        ListNode dummy = new ListNode(-1,head);
        // 翻转链表
        ListNode pre_first = dummy;
        ListNode pre_second = dummy;
        ListNode cur_first = dummy;
        ListNode cur_second = dummy;
        for(int i=1;i<left&&pre_first!=null;i++){
            pre_first = pre_first.next;
        }
        cur_first = pre_first.next;
        for(int i=1;i<=right&&cur_second!=null;i++){
            cur_second = cur_second.next;
        }
        pre_second = cur_second.next;
        // 翻转
        ListNode newHead = reverse(cur_first,pre_second);
        // 连接上
        pre_first.next = newHead;
        cur_first.next = pre_second;
        return dummy.next;
    }

    public ListNode reverse(ListNode head,ListNode tail){
        ListNode pre = null;
        ListNode next = null;
        while(head!=tail){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}
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