## LeetCode 21 合并链表 ## 引用官方代码,定义结点类 class ListNode: def __init__(self, x): self.val = x self.next = None 将数组转换为链表,也可以理解为链表的创建(官网不给出): ## 将给出的数组,转换为链表 def linkedlist(list): head = ListNode(list[0
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li
next = list1; } else { // 如果 list1 没有结点,表明 list2 已遍历完成, // 则将 list2 直接放在 tail 后面 tail.next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/...
*/functionmergeTwoLists($l1,$l2){// 设置一个哨兵$node=newListNode(null);$pre=$node;while($l1!=null&&$l2!=null) {if($l1->val <$l2->val) {$pre->next =$l1;$l1=$l1->next; }else{$pre->next =$l2;$l2=$l2->next;
本题中给定两个有序链表让我们进行重新连接,这里我们可以将其看作对两条链表重组的过程,而由于是有序链表,其连接必定是从左向右进行,我们设定两个指针,一个指针做头节点,另一个用来做追踪最小元素,在两个链表的遍历过程中先一一对比,较小的元素处链表指针后移一位,追踪指针指向该较小的节点,另一条链表的链表指...
https://leetcode.com/problems/merge-two-sorted-lists/ 我的思路就是找到第一个j > i的,然后用pre_j串起来。然后相应地对i再做一次。这种思路不太好。 note在指针前进的时候,判断越界的语句要放在if 或者while的前半部分。 my code: class Solution(object): ...
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 1. 2. # Definition for singly-linked list. ...
力扣Leetcode 21|合并两个有序链表Merge Two sorted Array 2020年10月13日 12:00595浏览·3喜欢·0评论 爱学习的饲养员 粉丝:6.7万文章:46 关注 视频讲解 622:17 Leetcode力扣 1-300题视频讲解合集|手画图解版+代码【持续更新ing】 77.6万783 视频爱学习的饲养员 ...
class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: head = ListNode(0) first = head while l1 and l2: if l1.val > l2.val: head.next = l2 l2 = l2.next else: head.next =...