9while(rightlist!=null&&leftlist!=null){ 10if(rightlist.val<leftlist.val){ 11ptr.next = rightlist; 12ptr = ptr.next; 13rightlist = rightlist.next; 14}else{ 15ptr.next = leftlist; 16ptr = ptr.next; 17leftlist = leftlist.next; 18} 19} 20 21if(rightlist!=null) 22ptr.next ...
Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题解: Merge k sorted linked list就是merge 2 sorted linked list的变形题。 而且我们很自然的就想到了经典的Merge Sort,只不过那个是对数组进行sort。而不同的地方,仅仅是Merge两个list的操作不同。 这里来...
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 1. 2. 3. 4. 5. 6. 代码 /** * Definition for singly-linked li...
// Importing the java.util package to use LinkedList, Scanner, ArrayList, and List import java.util.*; // Defining the Main class class Main { // Main method, the entry point of the program public static void main(String[] args) { // Creating the 1st linked list LinkedList list1 = ...
Sort a linked list using insertion sort. 1.解题思路 题目很简单,就是要求用插入排序的方法来为链表排序。插入排序就是每次遍历一个新的元素,将其插入到前面已经排好序的元素中。 因为头结点没法确定,所以我们用一个dummy节点;然后开始向后逐个遍历,当前遍历的节点为cur,另外sortnode每次都指向已经排好序的第一...
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. 题目要求我们将两个有序链表合成一个有序的链表。 Example: 输入: 1->2->4, 1->3->4 ...
Java Code:import java.util.* public class Solution { public static void main(String[] args) { // Create two sorted linked lists ListNode list1 = new ListNode(1); list1.next = new ListNode(3); list1.next.next = new ListNode(7); list1.next.next.next = new ListNode(9); list1....
next = l2; l2 = l2.next; } prev = prev.next; } // 合并后 l1 和 l2 最多只有一个还未被合并完,我们直接将链表末尾指向未合并完的链表即可 prev.next = l1 == null ? l2 : l1; return prehead.next; } // Definition for singly-linked list. public class ListNode { int val; ...
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) { 2. null, l, p = l1, q = l2; ...
Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 题目说明需要合并的列表已经排序过了,这点很重要。如果没有排序就复杂多了。 # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x ...