# 23.Merge k Sorted Lists (Array; Sort)

Source: Internet
Author: User

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

Idea I: Select sort

Each time you compare the Val of the head pointer of each list, take the smallest one. Time complexity O (n2)

Result:time Limit exceeded

Idea II: Minimum heap.

Time Complexity: Heap sequencing is also a sort of selection. Just select the sort directly, in order to select the maximum record from R[1...N], you need to compare n-1 times, then select the maximum record from r[1...n-2] to compare n-2 times. In fact, many of these n-2 comparisons have already been made in the previous n-1 comparisons, and the tree-shaped selection sort happens to preserve some of the previous comparisons by using the tree shape feature, thus reducing the number of comparisons. Time Complexity of O (NLOGN)

Heap sorting thought reference: http://jingyan.baidu.com/article/5225f26b057d5de6fa0908f3.html

`classSolution { Public: ListNode*mergeklists (Vector<listnode *> &lists) {        //using heap sorting,//1. Select the head of each linked list to insert into the small top heap,//2. The top of the heap is connected to the linked list,//3. The selected pointer moves back and then joins the small top heap, returning to 2//4. When all of the last linked lists are empty, return the header pointer of the merged list        if(Lists.empty ())returnnullptr; Vector<listnode* >Heap; Heap.push_back (0);//padding//1. Select the head of each linked list to insert into the small top heap,         for(inti =0; I! = Lists.size (); i + +){           if(Lists[i]) heap.push_back (lists[i]);        } makeheap (heap); //2. The top of the heap is connected to the linked list,ListNode Head (-1);//table header for merging listslistnode* p = &Head;  while(Heap.size () >1) {Auto Minnode= heap[1]; P->next = Minnode;//Access linked listp = p->Next; //3. The selected pointer moves back and then joins the small top heap, returning to 2Auto Next = minnode->Next; if(next) {heap[1] =Next; }Else{swap (heap[1], heap[heap.size ()-1]);            Heap.pop_back (); } minheap (Heap,1);//after adding a new element to the top of the heap, adjust it from top to bottom        }        //4. When all of the last linked lists are empty, return the header pointer of the merged list        returnHead.next; }    //build a small top pile//from bottom up    voidMakeheap (Vector<listnode*> &heap) {        //Create a small top heap starting from the parent of the last element         for(inti = (heap.size ()-1)/2; i >0; I--) {minheap (heap, i); }    }    //small Top heap, with the first element as the root to build a small top heap//position starting from 1, when taking elements remember-1//from top to bottom    voidMinheap (vector<listnode*> &heap,inti) {        intL = i*2; intR = L +1; intleast (i); //figure out the position of the smallest element        if((l< heap.size ()) && heap[l]->valval) {            //If the border is not exceeded and the left child is smaller than the father, changeleast =l; }        if(Rval) {            //if the boundary is not exceeded and the right child is the smallest, changeleast =R; }        if(Least! =i)            {Swap (Heap[i], heap[least]); Minheap (heap, least);//after the change, continue to adjust downwards        }    }};`

Idea III: Merge sort

Initial state: 6,202,100,301,38,8,1 after first merge: {6,202},{100,301},{8,38},{1}, number of comparisons: 3; After the second merge: {6,100,202,301},{1,8,38}, Comparison number: 4 ; After the third merge: {1,6,8,38,100,202,301}, compare number of times: 4;reference:http://baike.baidu.com/link?url=ayx3mqx_ Crmcjoxkl7ekhxuklh9pjkjsd1xdmap6eqwvffc-btnqbutselrafxbxqhcfoilkc5vsl14lgjejik Merge sort time complexity O (NLOGN)
`/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Mergeklists (vector<listnode*>&lists) {        if(Lists.empty ())returnNULL; MergeSort (lists,0, Lists.size ()-1); returnlists[0]; }    voidMergeSort (vector<listnode*>& lists,intStartintend) {        if(Start==end)return; intMid = (start + end) >>1;        MergeSort (lists, start, mid); MergeSort (lists, mid+1, end); Merge (Lists,start, Mid+1); }        voidMerge (vector<listnode*>& lists,intLst1,intlst2) {ListNode* Root =NULL; ListNode* current =NULL;  while(Lists[lst1] &&Lists[lst2]) {            if(Lists[lst1]->val <= lists[lst2]->val) {                if(!root) {Root=Lists[lst1]; Current=Root; }                Else{ Current->next =Lists[lst1]; Current= current->Next; } Lists[lst1]= lists[lst1]->Next; }            Else{                if(!root) {Root=Lists[lst2]; Current=Root; }                Else{ Current->next =Lists[lst2]; Current= current->Next; } Lists[lst2]= lists[lst2]->Next; }        }                 while(Lists[lst1]) {if(!root) {Root=Lists[lst1]; Current=Root; }            Else{ Current->next =Lists[lst1]; Current= current->Next; } Lists[lst1]= lists[lst1]->Next; }         while(Lists[lst2]) {if(!root) {Root=Lists[lst2]; Current=Root; }            Else{ Current->next =Lists[lst2]; Current= current->Next; } Lists[lst2]= lists[lst2]->Next; } Lists[lst1]=Root; }};`

23.Merge k Sorted Lists (Array; Sort)

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