Merge Two Sorted Lists & Remove Nth Node From End of List
1.合并两个排好序的list
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.
2.删除list倒数第n个元素
Remove Nth Node From End of List
Given a linked list, remove the nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
两个问题合在一个工程里面测试
package com.rust.datastruct;
public class MergeTwoSortedLists {
public static ListNode mergeTwoSortedLists(ListNode l1, ListNode l2) {
ListNode r1 = new ListNode(0);
ListNode res = r1;
ListNode t1 = l1;//java 中这样的赋值操作,对了l1操作等同于对t1操作
ListNode t2 = l2;
while (t1 != null && t2 != null){
if (t1.val <= t2.val) {
r1.next = t1;
t1 = t1.next;
} else {
r1.next = t2;
t2 = t2.next;
}
r1 = r1.next;
}
if (t1 != null) {
r1.next = t1;
}
if (t2 != null) {
r1.next = t2;
}
res = res.next;
return res;
}
/**
* @param head
* @param n
* @return ListNode
*/
public static ListNode removeNthFromEnd(ListNode head, int n) {
if (n == 0) return head;
ListNode fakeNode = head;
int count = 0;
while (fakeNode != null) {
fakeNode = fakeNode.next;
count++;
}
fakeNode = head;
if (n == count) {
head = head.next;
return head;
} else {
for (int i = 0; i < count; i++) {
if (i + n + 1== count) {
System.out.println(fakeNode.val);
ListNode cut = fakeNode.next.next;
fakeNode.next = cut;
count--;
continue;
}
fakeNode = fakeNode.next;
}
}
return head;
}
public static void main(String args[]){
ListNode l1 = new ListNode(0);
ListNode l2 = new ListNode(1);
ListNode p1 = l1;
ListNode p2 = l2;
/*initial the list*/
for (int i = 2; i <= 10; i++) {
if (i%2 == 0) {
p1.next = new ListNode(i);
p1 = p1.next;
} else {
p2.next = new ListNode(i);
p2 = p2.next;
}
}
p1 = l1;
p2 = l2;
System.out.println("input List l1 and l2");
showData(l1, l2);//after show,l1 and l2 value didn't change!
System.out.println("mergeTwoLists(l1, l2) -->");
ListNode res = mergeTwoSortedLists(l1, l2);
/** test mergeTwoSortedLists start ************/
while (res.next != null) {// res is destroyed
System.out.print(res.val + "\t");
res = res.next;
}
System.out.println(res.val);
System.out.println("After merge");
/** End ***********************************/
/** test removeNthFromEnd start **************/
showData(l1, l2);
// use l2 to test
int n = 1;
l2 = removeNthFromEnd(l2,n);
showData(l1, l2);
/** End ***********************************/
}
/**
* Print the ListNode
* @param l1 ListNode
* @param l2 ListNode
*/
public static void showData(ListNode l1, ListNode l2){
System.out.println("l1 -->");
while (l1.next != null) {
System.out.print(l1.val + "\t");
l1 = l1.next;
}
System.out.println(l1.val);
System.out.println("l2 -->");
while (l2.next != null) {
System.out.print(l2.val + "\t");
l2 = l2.next;
}
System.out.println(l2.val);
System.out.println("/************************************/");
}
}
//Definition for singly-linked list.
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
输出:
input List l1 and l2
l1 -->
0 2 4 6 8 10
l2 -->
1 3 5 7 9
/************************************/
mergeTwoLists(l1, l2) -->
0 1 2 3 4 5 6 7 8 9 10
After merge
l1 -->
0 1 2 3 4 5 6 7 8 9 10
l2 -->
1 2 3 4 5 6 7 8 9 10
/************************************/
9
l1 -->
0 1 2 3 4 5 6 7 8 9
l2 -->
1 2 3 4 5 6 7 8 9
/************************************/
Merge Two Sorted Lists & Remove Nth Node From End of List的更多相关文章
- [LeetCode] Merge k Sorted Lists 合并k个有序链表
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 这 ...
- Merge Two Sorted Lists
Merge Two Sorted Lists https://leetcode.com/problems/merge-two-sorted-lists/ Merge two sorted linked ...
- [Leetcode][Python]21: Merge Two Sorted Lists
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 21: Merge Two Sorted Listshttps://oj.le ...
- [Leetcode][Python]23: Merge k Sorted Lists
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 23: Merge k Sorted Listshttps://oj.leet ...
- LeetCode之“链表”:Merge Two Sorted Lists && Merge k Sorted Lists
1. Merge Two Sorted Lists 题目链接 题目要求: Merge two sorted linked lists and return it as a new list. The ...
- python 中的堆 (heapq 模块)应用:Merge K Sorted Lists
堆是计算机科学中一类特殊的数据结构的统称.堆通常是一个可以被看做一棵树的数组对象.在队列中,调度程序反复提取队列中第一个作业并运行,因为实际情况中某些时间较短的任务将等待很长时间才能结束,或者某些不短 ...
- 蜗牛慢慢爬 LeetCode 23. Merge k Sorted Lists [Difficulty: Hard]
题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity ...
- LeetCode: Merge k Sorted Lists 解题报告
Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and descr ...
- [Leetcode] Merge k sorted lists 合并k个已排序的链表
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 思 ...
随机推荐
- JavaSE教程-04Java中循环语句for,while,do···while
** Java的循环语句 ** 引入: 生活中有循环,程序的世界也有循环. 思考:生活中有哪些循环的事情? 总结:什么是循环? 重复做类似的事情,而且有终止条件,如果没有终止条件会是怎样? 类似这样的 ...
- css解决IE6,IE7,firefox兼容性问题
在当前IE6 IE7 FifeFox三分天下的情况下,浏览器兼容性问题成为网页设计者需要考虑的问题. 区别不同浏览器的CSS hack写法: 区别IE6与FF: background:orange;* ...
- Hql没有limit,替换方案
在HQL+MYSQL中不能直接在查询语句中使用LIMIT进行检索,正确方法为: String hql = "from User where id=? order by addDate des ...
- HTML 简单了解
HTML 特别的通俗易懂!想学自己制作网页的,就来我这看看吧! 首先 我先介绍一下什么是HTML! HTML是用来描述网页的一种语言!他结合CSS样式之后会有非常炫酷的样式! 1.HTML是指一种超文 ...
- Dalvik详解
整理自百度百科 Dalvik Dalvik是Google公司自己设计用于Android平台的Java虚拟机.Dalvik虚拟机是Google等厂商合作开发的Android移动设备平台的核心组成部分之一 ...
- Realm的简单使用
Realm个人理解: 它主要是一套移动端数据库框架,将对数据库表的操作转换为对对象的操作,如果是学过Java ORM框架的同学使用Realm起来应该上手比较快一点.而且据我所知Realm不仅支持Obj ...
- 添加OpenStack Mitaka源
sudo apt-get install ubuntu-cloud-keyring sudo add-apt-repository cloud-archive:mitaka 同理,可以添加其它版本,如 ...
- Ubuntu发行版升级
从UK 13.10升级到UK 14.10 方法一: 1.sudo apt-get update 2.sudo update-manager -c -d 3.选择upgrade(升级) 方法二 ...
- Testlink研究小结
1.Redmine与Testlink的关联 (1)redmine中的项目对应testlink的测试项目 (2)testllink执行用例时发现的问题可以记录到redmine中 2.Testlink优点 ...
- 【LCA求最近公共祖先+vector构图】Distance Queries
Distance Queries 时间限制: 1 Sec 内存限制: 128 MB 题目描述 约翰的奶牛们拒绝跑他的马拉松,因为她们悠闲的生活不能承受他选择的长长的赛道.因此他决心找一条更合理的赛道 ...