[路飞]leetcode-92. 反转链表 II

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题目描述

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]

提示: 链表中节点数目为 n 1 <= n <= 500 -500 <= Node.val <= 500 1 <= left <= right <= n

1、创建虚拟节点防止头节点丢失

2、查找到left-1节点,以此节点为遍历启始节点

3、具体反转思路文字解释不清楚直接上图

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var reverseBetween = function (head, left, right) {
  if (!head || !head.next) return head;
  let emptyNode = new ListNode(-1);
  emptyNode.next = head;
  let pre = emptyNode;
  for (let index = 0; index < left - 1; index++) {
    pre = pre.next;
  }
  let cur = pre.next; 
  let next = cur.next; 
  for (let index = 0; index < right - left; index++) {
    next = cur.next; 
    cur.next = next.next;
    next.next = pre.next;
    pre.next = next;
  }
  return emptyNode.next;
};