剑指offer 06 (opens new window)

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

示例 1：

输入：head = [1,3,2]
输出：[2,3,1]

限制：

• 0 <= 链表长度 <= 10000

解答：

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {number[]}
 */
var reversePrint = function(head) {
    if(!head) return [];
    if(!head.next) return [head.val];
    let res = [];
    let pre = null;
    let cur = head;
    while(cur) {
        let next = cur.next;
        cur.next = pre;
        pre = cur;
        cur = next;
    }
    // 翻转链表后再逐一输出
    while(pre) {
        res.push(pre.val);
        pre = pre.next;
    }
    return res;
};