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...
## 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...
Quest: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. 合并两个有序数列 题目给出的原型类 publicclassListNode {intval; ListNode next; ListNode(intx) { val =x; } }publicListNode mergeTwoLists(L...
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...
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...
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...
文章作者:Tyan 博客:noahsnail.com|CSDN|简书 1. Description Merge Two Sorted Lists 2. Solution /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} ...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
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
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 = ...