206.反转链表🔖链表

206. 反转链表

• https://leetcode-cn.com/problems/reverse-linked-list/

问题描述

问题思路

1. 利用外部空间：将所给链表存到ArryList里面或者是新的链表里面，然后再反转动态数组就可以了。
2. 快慢指针
3. 递归解法

代码实现

js

/**

- Definition for singly-linked list.
- function ListNode(val, next) {
-     this.val = (val===undefined ? 0 : val)
-     this.next = (next===undefined ? null : next)
- }
  \*/
  /\*\*
- @param {ListNode} head
- @return {ListNode}
  \*/
  var reverseList = function(head) {
  let prev = null;
  let curr = head;
  while (curr) {
  const next = curr.next;
  curr.next = prev;
  prev = curr;
  curr = next;
  }
  return prev;
  };

递归实现

/\*\*
- public class ListNode {
-     int val;
-     ListNode next;
-     ListNode(int x) { val = x; }
- }
  \*/

// 避免陷入死循环
  if (head == null || head.next == null) return head;
  ListNode newHead = reverseList(head.next); //此处递归，找到最后一个节点了
  head.next.next = head; //重新指定节点指向（有两个 next，注意少写）
  head.next = null; //将最初的节点指向空
  return newHead; //返回新的“倒置”头节点

快慢指针

class Solution {
  public ListNode reverseList(ListNode head) {
  // 避免陷入死循环
  if (head == null || head.next == null) return head;

          ListNode newHead = null;
          while (head != null){
              ListNode tmp = head.next;
              head.next = newHead;
              newHead = head;
              head = tmp;
          }
          return newHead;

      }

  }