public void addAfter(T element, T target) { if (size() == list.length) expandCapacity(); int scan = 0; while (scan < rear && !target.equals(list[scan])) scan++; if (scan == rear) throw new ElementNotFoundExcept
Write a Java program to merge the two sorted linked lists. Sample Solution: Java Code: importjava.util.*publicclassSolution{publicstaticvoidmain(String[]args){// Create two sorted linked listsListNodelist1=newListNode(1);list1.next=newListNode(3);list1.next.next=newListNode(7);list1.next....
详细代码注解见下。 递归解法(Java) /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classSolution {publicListNode deleteDuplicates(ListNode head) {if(head ==null|| head.next ==null)returnhead;//特殊...
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...
java sort 数字集合 java sorted list,一、stream基础方法使用1.filter过滤filter方法用于通过设置条件过滤出满足条件的元素。以下代码片段使用filter方法过滤出空字符串。List<String>list=Arrays.asList("abc","","bc","efg","abcd","","jkl");List<String>f
从有序链表中删除重复的数字,并且返回删除后的头结点 例如输入链表为1->1->2,返回1->2 这题和leetcode26相似,只是数据结构从数组变成了链表 /** * @author rale * * Given a sorted linked list, delete all duplicates such that each element appear only once. ...
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. 依次拼接 复杂度 时间O(N) 空间 O(1) 思路 该题就是简单的把两个链表的节点拼接起来,我们可以用一个Dummy头,将比较过后的节点接在这个Dummy头之后。
代码: java: 代码语言:javascript 代码运行次数:0 运行 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution{publicListNodedeleteDuplicates(ListNode head){if(head==null||head.next==null)return...
java: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution{publicListNodedeleteDuplicates(ListNode head){if(head==null||head.next==null)...
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 # self.next = None class Solution: ...