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 new list should be made by splicing together the nodes of the first two lists.

  这道题目题意是要将两个有序的链表合并为一个有序链表。为了编程方便,在程序中引入dummy节点。具体程序如下:

 /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode *dummy = new ListNode();
ListNode *head = nullptr, *start = dummy;
int flag = true;
while(l1 && l2)
{
if(l1->val < l2->val)
{
start->next = l1;
l1 = l1->next;
}
else
{
start->next = l2;
l2 = l2->next;
}
start = start->next;
} if(!l1)
start->next = l2;
else if(!l2)
start->next = l1; head = dummy->next;
delete dummy;
dummy = nullptr; return head;
}
};

  2. Merge k Sorted Lists

   题目链接

  题目要求:

  Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

  这道题本来想用比较直观的方法,但超时了。具体程序如下:

 /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
int sz = lists.size();
if(sz == )
return nullptr; ListNode *node = new ListNode();
ListNode *head = nullptr, *start = node;
while(true)
{
int count = ;
int minVal = INT_MAX;
int minNode = -;
for(int i = ; i < sz; i++)
{
if(lists[i] && lists[i]->val < minVal)
{
minNode = i;
minVal = lists[i]->val;
}
else if(!lists[i])
count++;
} if(count != sz)
{
start->next = lists[minNode];
lists[minNode] = lists[minNode]->next;
}
else
{
head = node->next;
delete node;
node = nullptr; return head;
} start = start->next;
}
}
};

  后来参考网上的做法,用Divide and Conquer方法会比较好,具体程序如下:

 /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode *node = new ListNode();
ListNode *head = nullptr, *start = node;
int flag = true;
while(l1 && l2)
{
if(l1->val < l2->val)
{
start->next = l1;
l1 = l1->next;
}
else
{
start->next = l2;
l2 = l2->next;
}
start = start->next;
} if(!l1)
start->next = l2;
else if(!l2)
start->next = l1; head = node->next;
delete node;
node = nullptr; return head;
} ListNode *mergeKLists(vector<ListNode*>& lists, int low, int high)
{
if(low < high)
{
int mid = (low + high) / ;
return mergeTwoLists(mergeKLists(lists, low, mid), mergeKLists(lists, mid + , high));
}
return lists[low];
} ListNode* mergeKLists(vector<ListNode*>& lists) {
int sz = lists.size();
if(sz == )
return nullptr; return mergeKLists(lists, , sz - );
}
};

  算法分析摘自同一博文

  我们来分析一下上述算法的时间复杂度。假设总共有k个list,每个list的最大长度是n,那么运行时间满足递推式T(k) = 2T(k/2)+O(n*k)。根据主定理,可以算出算法的总复杂度是O(nklogk)。如果不了解主定理的朋友,可以参见主定理-维基百科。空间复杂度的话是递归栈的大小O(logk)。

LeetCode之“链表”:Merge Two Sorted Lists && Merge k Sorted Lists的更多相关文章

  1. [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 ...

  2. [LeetCode] 23. 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. E ...

  3. 【LeetCode】23. Merge k Sorted Lists 合并K个升序链表

    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:合并,链表,单链表,题解,leetcode, 力扣,Py ...

  4. 【leetcode】Merge k Sorted Lists(按大小顺序连接k个链表)

    题目:Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity ...

  5. [leetcode]23. Merge k Sorted Lists归并k个有序链表

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. I ...

  6. 【LeetCode每天一题】 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. E ...

  7. LeetCode Merge k Sorted Lists (链表)

    题意 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity ...

  8. LeetCode OJ:Merge k Sorted Lists(归并k个链表)

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 类 ...

  9. [LeetCode]23. Merge k Sorted Lists合并K个排序链表

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. E ...

随机推荐

  1. Tomcat7源码环境搭建

    一.下载Tomcat7源码 从官网上下载Tomcat源码,   http://mirror.bit.edu.cn/apache/tomcat/tomcat-7/v7.0.70/src/apache-t ...

  2. Kafka系列之-自定义Producer

    前面已经讲到了,在Kafka中,Message是由Producer产生的,Producer产生的Message会发送到Topic的指定Partition中.Producer可以有多种形式,也可以由用户 ...

  3. linux 防火墙操作

    root/12345 (只能用ROOT操作)iptables -I INPUT -s x.x.x.x -p tcp --dport 8091 -j ACCEPT   #允许x.x.x.x访问本机的80 ...

  4. Scala:提取器(Extractor)

    http://blog.csdn.net/pipisorry/article/details/52902671 提取器是从传递给它的对象中提取出构造该对象的参数. Scala 标准库包含了一些预定义的 ...

  5. 传Lua对象到Cpp

    传Lua对象到Cpp (金庆的专栏) 摘自:http://raycast.net/lua-intf 以下代码演示了Lua函数和表传入Cpp进行处理: std::string acceptStuff(L ...

  6. pandas小记:pandas数据输入输出

    http://blog.csdn.net/pipisorry/article/details/52208727 数据输入输出 数据pickling pandas数据pickling比保存和读取csv文 ...

  7. Android自定义处理崩溃异常

    用过安卓手机的用户以及安卓开发者们会时长碰到程序异常退出的情况,普通用户遇到这种情况,肯定非常恼火,甚至会骂一生垃圾软件,然后卸载掉.那么开发者们在开发过程中遇到这种情况给怎么办呢,当然,你不可能世界 ...

  8. Spark技术内幕: Task向Executor提交的源码解析

    在上文<Spark技术内幕:Stage划分及提交源码分析>中,我们分析了Stage的生成和提交.但是Stage的提交,只是DAGScheduler完成了对DAG的划分,生成了一个计算拓扑, ...

  9. GCD API记录(二)

    前言 这是关于GCD的第二篇文章,GCD的API有100多个,通过快捷键Option + 单击,可以在Reference中的Grand Central Dispatch (GCD) Reference ...

  10. java中遍历map的几种方法介绍

          喜欢用Java写程序的朋友都知道,我们常用的一种数据结构map中存储的是键值对,我们一般存储的方式是: map.put(key, value); 而提取相应键的值用的方法是: map.ge ...