## 解法二:递归 Recursion class Solution: def mergeTwoLists(self, head1, head2): ## head1 和 head2 头结点指代两个链表 ## 递归的 base,结束条件 if head1 is None: return head2 elif head2 is None: return head1 elif head1.val < head2.val: ## 把小的值 head1 提取出来,拼接上后面...
2.Solutions: 1/**2* Created by sheepcore on 2019-05-103* Definition for singly-linked list.4* public class ListNode {5* int val;6* ListNode next;7* ListNode(int x) { val = x; }8* }9*/10classSolution {11publicListNode mergeTwoLists(ListNode l1, ListNode l2) {12ListNode head =n...
3.Java代码: (1)非递归: 为方便操作,定义一个辅助的头节点,然后比较原来两个链表的头节点,将小的那一个加入到合并链表,最后,当其中一个链表为空时,直接将另一个链表接入到合并链表即可。 1//public class LeetCode21 为测试2publicclassLeetCode21 {3publicstaticvoidmain(String[] args) {4ListNode m1=new...
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. 3. 4. 5. 6. 代码 /** * Definition for singly-linked li...
1. Merge Two Sorted Lists 题目链接 题目要求: 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. 这道题目题意是要将两个有序的链表合并为一个有序链表。为了编程方便,在程序中引入dummy节点。具体程序...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
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/problem LeetCode 日更第 52 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
二、代码实现 # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = NoneclassSolution(object):defmergeTwoLists(self,l1,l2):""" :type l1: ListNode :type l2: ListNode ...
Leetcode 21 Merge Two Sorted Lists 题目详情 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. 题目要求我们将两个有序链表合成一个有序的链表。
class Solution { public: void heapify(vector<int> &arr, int i, int n) { int largest = i; int left = 2 * i + 1; int right = 2 * i + 2; if (left < n && arr[left] > arr[largest]) { largest = left; } if (right < n && arr[right] > arr[largest]) { largest = ...