206. 反转链表

[webVueBlog]

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。   示例 1： 输入：head = [1,2,3,4,5] 输出：[5,4,3,2,1]

示例 2： 输入：head = [1,2] 输出：[2,1]

示例 3： 输入：head = [] 输出：[]   提示：

链表中节点的数目范围是 [0, 5000] -5000 <= Node.val <= 5000

/**
 * 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}
 206. 反转链表
 迭代的方法

      pre = null
      cur = head
(指向空节点)
 pre  cur(指向当前的头节点)
 null  1 -> 2 -> 3 -> 4 -> 5 -> null
 pre = null
 cur = head
 当当前节点不为空的时候

 pre    cur  next
 null <- 1   2     ->  3 -> 4 -> 5 -> null
 next = current.next
 cur.next = pre

            cur
       pre  next
null <- 1    2   ->  3 -> 4 -> 5 -> null
 pre = cur
 cur = next
 next = current.next
 cur.next = pre
 
 直到cur节点为空 return pre
 */
var reverseList = function(head) {
    let pre = null, cur = head, next
    // 判断当前指针
    while(cur) {
        next = cur.next
        cur.next = pre
        pre = cur
        cur = next
    }
    return pre
};

[webVueBlog]

/**
 * 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}
 206. 反转链表
递归的方法


 */
var reverseList = function(head) {
    if(head === null || head.next === null) {
        return head
    }
    const node = reverseList(head.next)
    head.next.next = head
    head.next = null
    return node
};