https://leetcode.cn/problems/reverse-linked-list/ (opens new window)

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:

        pre = None
        cur = head
        while cur:
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next

        return pre

/**
 * 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 cur = head
    let pre = null
    while(cur != null){
        const t = cur.next
        cur.next = pre
        pre = cur
        cur = t
    }
    return pre
};

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
//  参考leetcode cn 官网 解答
var reverseList = function(head) {
    if(head == null || head.next == null) return head
    
    let newHead = reverseList(head.next)
    // 前一个节点是有后一个节点的信息， 
     // 然后前一个节点->后一个节点， 转变成,  后一个节点 -> 前一个节点
    head.next.next = head

    // 前一个节点的next 需要变成 null
    head.next = null
    return newHead
};