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. 分析:合并两个有序序列,这个归并排序中的一个关键步骤.这里是要合并两个有序的单链表.由于链表的特殊性,在合并时只需要常量的空间复杂度. 编码: public ListNode mergeTwoLists(Li…

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 这道混合插入有序链表和我之前那篇混合插入有序数组非常的相…

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. 这道混合插入有序链表和我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.具体思想就是新建一个链表,然后比较两个链表中的元素值,把较小的…

Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splicing together the nodes of the two lists and sorted in ascending order. Have you met this question in a real interview? Yes Exampl…

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. 题目标签:Linked List 题目给了我们两个lists,让我们有序的合并两个 lists. 这题利用递归可以从list 的最后开始向前链接nodes,代码很简洁,清楚. Java Solution: Runti…

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. Example:  Input:1->2->4, 1->3->4  Output:1->1->2->3->4->4 思路 思路一:建立一个新链表 建立一个新的链表,用…

题意:合并两个有序链表 /** * 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) { if(l1 == NULL) retu…

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. 解决方案:276ms /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next…

21. Merge Two Sorted Lists 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 需要排序!!! [2,4][1] 输出1 2 4 而不是2 4 1  递归版!! class Solution { public ListNode mergeTwo…

21.Merge Two Sorted Lists 初始化一个指针作为开头,然后返回这个指针的next class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = ); ListNode* p = dummy; while(l1 && l2){ if(l1->val <= l2->val){ p->next = l1; p = p-&…