力扣解题 206. 反转链表

解法一

/*
* @lc app=leetcode.cn id=206 lang=javascript
*
* [206] 反转链表
*/

// @lc code=start
/**
* 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(heaXQd) {
    /**
    *  输入：head = [1,2,3,4,5]
    *  输出：[5,4,3,2,1]
    * 1 2 3 4 5 
    * 5 4 3 2 1 
    */
    let prev = null, curr = head
    while (curr) {
        const next = curr.next // head的下一个节点
        curr.next = prev        // 让curr节点的next指向prev 
        prev = curr 
        curr = next
    }
    return prev

};
// @lc code=end

/*
* @lc app=leetcode.cn id=206 lang=java
*
* [206] 反转链表
*/

// @lc code=start
/**
* 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 reverseList(ListNode head) {
        // [1,2,3,4,5]
        // [5,4,3,2,1]
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next; // 首次进入循环时， curr为 1 ， curr.next 为2
            curr.next = prev; // 先将curr的指针指向prev，
            curr = next; // 然后next位置给curr, 
            prev = curr; // 然后curr位置给prev,
        }
        return prev;
    }
}
// @lc code=end

解法二 递归

/*
* @lc app=leetcode.cn id=206 lang=javascript
*
* [206] 反转链表
*/

// @lc code=start
/**
* 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(heaXQd) {
    /**
    *  输入：[1,2,3,4,5]
    *  输出：[5,4,3,2,1]
    */
    if(heaXQd == null || heaXQd.next == null){
        return heaXQd;
    }
    let p = reverseList(heaXQd.next);
    heaXQd.next.next = heaXQd;
    heaXQd.next = null;
    return p;
};
// @lc code=end

/*
* @lc app=leetcode.cn id=206 lang=java
*
* [206] 反转链表
*/

// @lc code=start
/**
* 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 reverseList(ListNode head) {
        // [1,2,3,4,5]
        // [5,4,3,2,1]
        if(head == null || head.next == null){
            return head;
        }
        // p 是最里面的节点。
        // 上半部分是往里递的过程
        ListNode p = reverseList(head.next);
        // 下半部分是往回归的过程
        // 将最后一个节点的指针 指向 倒数第二个，
        head.next.next = head;  // 当前节点的的下个节点的指针指向当前节点。
        // 然后将倒数第二个指针指向null
        head.next = null;
        return p;
    }
}
// @lc code=end