当list1 和 list2 均还有结点时,取它们中较小的头结点放入结果链表中,然后不断循环。 最后当其中一个链表为空时,将另一个链表剩余的部分全部插入结果链表尾部即可。 进阶: LeetCode 23: 合并 k 个有序链表 LeetCode 148: 对无序链表排序 时间复杂度:O(|list1| + |list2|) 需要遍历 list1 中的全部 ...
## LeetCode 21 合并链表 ## 引用官方代码,定义结点类 class ListNode: def __init__(self, x): self.val = x self.next = None 将数组转换为链表,也可以理解为链表的创建(官网不给出): ## 将给出的数组,转换为链表 def linkedlist(list): head = ListNode(list[0]) ## 列表的第一个元素。这里...
自上而下思考,递归就是将问题向下分解成若干相同子问题,直到子问题规模最小,无法继续分解,即base case,求得最小问题的解后,再依次向上归。 递归两个两个条件:1)子问题调用同一个递归函数 2)递归终止条件(递到最小子问题,返回,向上归) /*** Definition for singly-linked list. * public class ListNode { ...
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 Tip:将两个有序链表融合,按照val值大小从小到大排序。 解法一 使用新的链表头...
https://leetcode.com/problems/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: easy 。 算法: 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) {...
LeetCode-cn 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] ...
https://leetcode.com/problems/merge-two-sorted-lists/description/ 将两条有序的链表合并为一条有序的链表 时间复杂度0(m+n) m, n 为链表的长度 /** * Definition for singly-linked list. * public class ListNode { * public var val: Int ...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
2015-07-13 10:31 − https://leetcode.com/problems/merge-two-sorted-lists/ 1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ... 阿怪123 0 122 Merge Two Sorted Lists 2015-09-25 19:48 − Merge two sorted linked lists and return it...
res.add(new Interval(start, end)); //1. add prev to result list start = interval.start; //2. update prev bounds end = interval.end; } else end = Math.max(end, interval.end); //else just update prev end bound } res.add(new Interval(start, end)); //add the prev which was ...