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->4Output:1->1->2->3->4->4 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一...
[LeetCode] Merge K Sorted Linked Lists 题目链接: https://oj.leetcode.com/problems/merge-k-sorted-lists/ 问题: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解题思路: 方法一:暴力破解 思路: 将列表一个一个地合并(例如:lists = [l1, l2...
//leetcode.cn/problems/middle-of-the-linked-list/ """ class Solution1: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: # 迭代法 dummy = cur = ListNode(0) while list1 and list2: # 当有为空的时候可以停止了把剩下的接上去就行。
改进mergeTwoLists方法,以在开始时检查空链表。 class ListNode: def __init__(self, x): self.val = x self.next = None # 改进后的将给出的数组转换为链表的函数 def linkedlist(list): if not list: # 检查列表是否为空 return None # 空列表返回None head = ListNode(list[0]) cur = head for...
* Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode*mergeTwoLists(struct ListNode*l1,struct ListNode*l2){} 通过函数定义可以看出,mergeTwoLists 函数的参数是两个单向链表,然后返回值也是一个链表,我们要做的就是把两个有序的链表合...
l2->next = mergeTwoLists(l1, l2->next); 完整测试代码: #include <vector> #include <string> #include<list> #include<algorithm> #include <sstream> #include<iostream> using namespace std; /* * @lc app=leetcode.cn id=21 * * [21] 合并两个有序链表 ...
* Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } *//** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */constmergeTwoLists=function(l1,l2){if(l1===null){returnl2;}if(l2===null){returnl1;}if...
时间复杂度:O(n+m),其中 n 和 m 分别为两个链表的长度。因为每次调用递归都会去掉 l1 或者 l2 的头节点(直到至少有一个链表为空),函数 mergeTwoList 至多只会递归调用每个节点一次。因此,时间复杂度取决于合并后的链表长度,即O(n+m)。 空间复杂度:O(n+m),其中 n 和 m 分别为两个链表的长度。递归调...
Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1: Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] Example 2: ...
public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if(l1 == null){ return l2; } if(l2 == null){ return l1; } ListNode p1 = l1; ListNode p2 = l2; //l1作主链 if(l1.val <= l2.val){ while(p2 != null){ //开路指针,遍历l2 ...