## 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...
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...
*@returnListNode */functionmergeTwoLists($l1,$l2){if($l1==null) {return$l2; }if($l2==null) {return$l1; }if($l1->val <=$l2->val) {$l1->next =mergeTwoLists($l1->next,$l2);return$l1; }if($l2->val <=$l1->val) {$l2->next =mergeTwoLists($l2->next,$l1);return$l2;...
public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if(l1==NULL) return l2; if(l2==NULL) return l1; if(l1->val<l2->val){ l1->next=mergeTwoLists(l1->next,l2); return l1; } else{ l2->next=mergeTwoLists(l1,l2->next); return l2; } } }; 1. 2. 3. 4. 5. 6...
The number of nodes in both lists is in the range[0, 50]. -100 <= Node.val <= 100 Bothlist1andlist2are sorted innon-decreasingorder. 这道混合插入有序链表和博主之前那篇混合插入有序数组非常的相似Merge Sorted Array,仅仅是数据结构由数组换成了链表而已,代码写起来反而更简洁。具体思想就是新建...
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 ...
Merge Two Sorted Lists 二、解题 两个有序链表的排序。 三、尝试与结果 classSolution(object):defmergeTwoLists(self,l1,l2):ret=ListNode(0)ifl1==None:returnl2ifl2==None:returnl1ifl1.val<l2.val:ret=l1 ret.next=self.mergeTwoLists(l1.next,l2)else:ret=l2 ...
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 lists.Return the head of the merged linked list.impl Solution { pub fn me
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 示例 输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4 解题思路: 简单考察对链表的使用 首先先创建一个虚拟头结点head 因为l1,l2两个链表都已经是从小到大排序完成,所以只要判断当前两个链表位置的...
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. 翻译:合并两个排好序的链列,并将其作为新链表返回。新链表应通过将前两个列表的节点拼接在一起。