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;

    }
}