Merge k Sorted Lists Solution

Question

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

Solution 1--Merge Sort

We can follow the method of the merge sort. Time Complexity O (Nlog (n)), n is the total number of all nodes.

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode mergeklists (listnode[] lists) { One         if(Lists = =NULL|| Lists.length < 1) A             return NULL; -         if(Lists.length = = 1) -             returnLists[0]; the         intK =lists.length; -         returnMsort (lists, 0, k-1); -     } -      +     PrivateListNode Msort (listnode[] lists,intStartintend) { -         if(Start <end) { +             intMid = (End-start)/2 +start; AListNode left =msort (lists, start, mid); atListNode right = Msort (lists, Mid + 1, end); -             returnmergetwolists (left, right); -         } -         returnLists[start]; -     } -      in     PrivateListNode mergetwolists (ListNode A, ListNode b) { -ListNode Newhead =NewListNode (0); toListNode p = newhead, p1 = A, p2 =b; +          while(P1! =NULL&& P2! =NULL) { -             if(P1.val <p2.val) { theP.next =NewListNode (p1.val); *P1 =P1.next; $}Else {Panax NotoginsengP.next =NewListNode (p2.val); -P2 =P2.next; the             } +p =P.next; A         } the          while(P1! =NULL) { +P.next =NewListNode (p1.val); -P1 =P1.next; $p =P.next; $         } -          while(P2! =NULL) { -P.next =NewListNode (p2.val); theP2 =P2.next; -p =P.next;Wuyi         } the         returnNewhead.next; -     } Wu}

Solution 2--Priorityqueue

We can use the Priorityqueue in Java. The elements of the priority queue is ordered according to their natural ordering, or by a comparator provided at the con Struction time.

Priorityqueue

Comparator

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {val = x;}7  * }8  */9  Public classSolution {Ten      PublicListNode mergeklists (listnode[] lists) { One         if(Lists = =NULL|| Lists.length < 1) A             return NULL; -ListNode Newhead =NewListNode (0); -Priorityqueue<listnode> PQ =NewPriorityqueue<listnode>(Lists.length, the                                             NewComparator<listnode>() { -                                                  Public intCompare (ListNode A, ListNode b) { -                                                     return(A.val-b.val); -                                                 }}); +          for(ListNode tmp:lists) { -             if(TMP! =NULL) + Pq.add (TMP); A         } atListNode p =Newhead; -          while(Pq.size () > 0) { -ListNode tmp =Pq.poll (); -P.next =tmp; -             if(Tmp.next! =NULL) - Pq.add (tmp.next); inp =P.next; -         } to         returnNewhead.next; +     } -}

Merge k Sorted Lists Solution