LeetCode23 Merge k Sorted Lists connects K ordered linked lists into one

title :

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

translation :

The K-ordered list is synthesized and returned.

Ideas :

This problem is similar to yesterday's Merge 2 lists. But K is not sure, if you use each to go through, it certainly will not pass.

So, can think of is the merging method.

I also remember the first freshman group arranged the merger of the work, at that time just get started, see the foggy.

Just take this opportunity to merge a good review.

Let's talk about the solution to this problem first:

Because it is a list of lists, you can divide it into 2 groups, then merge 2Lists for each group of 2.

Then in 4 groups, because the previous step has formed 2 chains, so this can also be done through the merge 2Lists, recursion to get results.

Code :

     Public ListNode mergeklists (listnode[] lists) {  if (lists = = NULL | | lists.length = = 0)  return null;  Return mergesort (lists, 0, lists.length-1);   }  public static ListNode MergeSort (listnode[] lists,int l, int r)  {  if (l>=r)  return lists[l];  int m = (l+r)/2;  ListNode ll = mergesort (lists, L, m);  ListNode rr = mergesort (lists, m+1, R);  Return Merge (LL,RR);      }  public static ListNode Merge (ListNode l1,listnode L2)  {  ListNode root =new listnode (0);  ListNode pre = root;  Pre.next = L1;  while (true)  {  if (L1 = = null)  {  pre.next = L2;  break;  }  if (L2 = = null)  {  pre.next = L1;  break;  }  if (L1.val < L2.val)  {  pre.next = L1;  L1 = L1.next;  }  else  {  pre.next = L2;  L2 = L2.next;   }  Pre = Pre.next;    }  return root.next;    }

Well, the code for this problem is above.

Here, we still have to write a merger. Because after all, it will be used later.

Merge sort Wiki Explanation

The personal understanding is to divide a long into a small part, that is, divide and conquer, divided into 2 groups, sorting each group after it is done in 4 groups.

Finally, the whole ordered arrangement is obtained.

A picture of Baidu, estimated to be helpful to understanding

Explanation of Wiki :

Merge sort works as follows (assuming that the sequence has n elements):

1. Merges each contiguous two digits of a sequence, forming a sequence that contains two elements for each sequence after sorting
2. Merge the above sequence again, forming a sequence with four elements per sequence
3. Repeat step 2 until all elements are sorted

So, write a merge sort yourself.

public class MergeSort {<span style= "white-space:pre" ></span><span style= "White-space:pre" ></ span>static int Number=0;<span style= "white-space:pre" ></span>public static void MergeSort (int arr[], int l, int r) <span style= "White-space:pre" ></span>{<span style= "White-space:pre" ></span>if ( L >= R) <span style= "White-space:pre" ></span>return; <span style= "White-space:pre" ></span >int mid = (l+r)/2;<span style= "White-space:pre" ></span>mergesort (arr, L, mid);//split into groups <span style= " White-space:pre "></span>mergesort (arr, mid+1, R); <span style=" White-space:pre "></span>Merge ( Arr, L, Mid, R);//Intra-group sort <span style= "White-space:pre" ></span>}<span style= "White-space:pre" ></ Span>public static void Merge (int arr[],int l,int Mid, int r) <span style= "White-space:pre" ></span>{< Span style= "White-space:pre" ></span>int []tep = new INT[ARR.LENGTH];&LT;SPAn style= "White-space:pre" ></span>int mm = Mid+1;<span style= "White-space:pre" ></span>int count = L,ll = L;<span style= "White-space:pre" ></span>while (L <= mid && mm <= R) <span style= "White -space:pre "></span>{<span style=" White-space:pre "></span>if (Arr[l]  <= arr[mm]) < Span style= "White-space:pre" ></span>tep[count++]  = arr[l++];<span style= "White-space:pre" > </span>else<span style= "White-space:pre" ></span>tep[count++] = Arr[mm++];<span style= " White-space:pre "></span>}<span style=" White-space:pre "></span>while (l<= mid) <span Style= "White-space:pre" ></span>{<span style= "White-space:pre" ></span>tep[count++] = arr[l++] ; <span style= "White-space:pre" ></span>}<span style= "White-space:pre" ></span>while (mm < =R) <span style= "White-space:pre" ></span>{<span style= "White-space:pre" ></span>tep[count++] = Arr[mm++];<span style= "white-space:pre" ></span>}<span style= "White-space:pre" ></span> System.out.println ("++number" + "trip \ T") <span style= "White-space:pre" ></span> while (ll<=r) <span style= "White-space:pre" ></span>{<span style= "White-space:pre" ></span >   &NBSP;ARR[LL] = Tep[ll];<span style= "White-space:pre" ></span>    system.out.print ( ARR[LL] + "\ t"); <span style= "White-space:pre" ></span>    ll++;<span style= "White-space:pre" ></span>}<span style= "White-space:pre" ></span>    system.out.println (); <span Style= "White-space:pre" ></span><span style= "White-space:pre" ></span>}<span style= " White-space:pre "></span>/**<span style=" White-space:pre "></span> * @param args<span style=" White-space:pre "></span> */<span style=" White-space:pre "></span>public static void MaIn (string[] args) {<span style= "white-space:pre" ></span>//TODO Auto-generated method stub <span style= "White-space:pre "></span>int[] A = {9,20,32,12,53,23,54,2};<span style=" White-space:pre "></span>     MergeSort (A, 0, a.length-1); <span style= "White-space:pre" ></span>for (int t:a) <span style= " White-space:pre "></span>{<span style=" White-space:pre "></span>system.out.println (t);< Span style= "White-space:pre" ></span>}<span style= "White-space:pre" &GT;&LT;/SPAN&GT;}}

Output Result:

The effect of sorting after grouping can be clearly seen in the output.

After the submission, I saw a group of people write in C only more than 10 milliseconds passed. Sure enough, Dick!

We'll study how they are written. And then in the modification.

LeetCode23 Merge k Sorted Lists connects K ordered linked lists into one