【LeetCode算法-21】Merge Two Sorted Lists LeetCode第21题 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 翻译: 合并两...
【LeetCode】21. Merge Two Sorted Lists 题目: 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. 思路:比较每个列表的第一个元素。 合并小的添加到列表中。 最后,当其中一个是空的,只需将它附加到合并...
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. Seen this question in a real interview before? Yes 简单的归并排序,不说了,直接上代码:
LeetCode-Java-21. Merge Two Sorted Lists 题目 AI检测代码解析 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...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists. Example: Input:1->2->4,1->3->4Output:1->1->2->3->4->4 ...
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) {} * }; */class Solution{public:ListNode*mergeTwoLists(ListNode*l1,ListNode*l2){ListNode*ptrFirst=l1...
每日算法——leetcode系列问题 Merge Two Sorted Lists Difficulty: EasyMerge 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./** * Definition for singly-linked list. * struct ListNode { * int val; * ...
23. Merge k Sorted Lists Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example: Input:[ 1->4->5, 1->3->4, 2->6 ]Output:1->1->2->3->4->4->5->6 思路: 合并k个有序链表,采用分治的思想,时间复杂度O(nlogk) ...
next return list class Solution: def mergeTwoLists(self, head1, head2): if not head1: # 如果head1为空,直接返回head2 return head2 if not head2: # 如果head2为空,直接返回head1 return head1 pre = ListNode(0) # 使用哑结点简化操作 head = pre while head1 and head2: if head1.val ...
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/...