Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
提议很简单,就是归并排序。
首先想到的即使逐个归并得到最终的结果,但是会超时,这是因为这种会造成数组的size大小不一样,导致归并排序的时间变长;
最好的做法是两两合并,然后在两两合并,这样不会超时,
需要注意的是cap=(size+1)/2,这个是考虑到奇数的情况。
代码如下:
/* class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}*/
public class Solution
{
/*
* 按照次序两两合并,会超时
* */
public ListNode mergeKLists111(ListNode[] lists)
{
ListNode head=new ListNode(-1);
if(lists.length<=0)
return head.next;
if(lists.length==1)
return lists[0];
//这个循环属于遍历方法
for(int i=0; i<lists.length;i++)
head.next=merge(head.next,lists[i]);
return head.next;
}
/*
* 注意分析数组元素数量为0个和1个时候的特殊情况
*
* 很明显可以使用二分方法去做
* */
public ListNode mergeKLists(ListNode[] lists)
{
ListNode head=new ListNode(-1);
if(lists.length<=0)
return head.next;
if(lists.length==1)
return lists[0];
int len=lists.length;
while(len>1)
{
int cap=(len+1)/2;
for(int i=0;i<len/2;i++)
lists[i]=merge(lists[i], lists[i+cap]);
len=cap;
}
return lists[0];
}
private ListNode merge(ListNode listNode, ListNode listNode2)
{
ListNode iter1=listNode;
ListNode iter2=listNode2;
ListNode head=new ListNode(-1);
ListNode he=head;
while(iter1!=null && iter2!=null)
{
if(iter1.val<=iter2.val)
{
he.next=new ListNode(iter1.val);
he=he.next;
iter1=iter1.next;
}
else
{
he.next=new ListNode(iter2.val);
he=he.next;
iter2=iter2.next;
}
}
while(iter1!=null)
{
he.next=new ListNode(iter1.val);
he=he.next;
iter1=iter1.next;
}
while(iter2!=null)
{
he.next=new ListNode(iter2.val);
he=he.next;
iter2=iter2.next;
}
return head.next;
}
}
C++版本的答案如下,这个和CUDA的归并很像,
代码如下:
#include <iostream>
#include <vector>
using namespace std;
/*
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
*/
class Solution
{
public:
ListNode* mergeKLists(vector<ListNode*>& lists)
{
if (lists.size() <= 0)
return NULL;
if (lists.size() == 1)
return lists[0];
int len = lists.size();
while (len > 1)
{
int cap = (len + 1) / 2;
for (int i = 0; i < len/2; i++)
lists[i] = mergeTwoLists(lists[i], lists[i + cap]);
len = cap;
}
return lists[0];
}
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
ListNode* head = new ListNode(-1);
ListNode* a = head;
while (l1 != NULL && l2 != NULL)
{
if (l1->val < l2->val)
{
a->next = new ListNode(l1->val);
l1 = l1->next;
a = a->next;
}
else
{
a->next = new ListNode(l2->val);
l2 = l2->next;
a = a->next;
}
}
while (l1 != NULL)
{
a->next = new ListNode(l1->val);
l1 = l1->next;
a = a->next;
}
while (l2 != NULL)
{
a->next = new ListNode(l2->val);
l2 = l2->next;
a = a->next;
}
return head->next;
}
};