Leetcode 23. Merge k Sorted Lists

Merge ksorted linked lists and return it as one sorted list. Analyze and describe its complexity.This problem if there is a 21st merger of two linked list of the basis will be easier, the specific merger linked list when there are two ways
(1) If the K list is connected in turn (L1 and L2 together, the result and L3 together, in turn), the time complexity is n*k*k will time out, the specific calculation method is as follows:Suppose you have a list of k lengths of N (l1,l2,l3...lk), and then calculate the worst-case time complexity,L1+L2---> l12 worst-case operation is 2N timesL12+L3---> l123 worst-case operation is 3N timesAnd so on : Worst case total operation: 2n+3n+4n+...+kn = (2+k) * (k-1) *N/2the time complexity of N*k*k
(2) If the 22 is connected, like a merge sort, the time complexity is N*K*LOGK will not time out;Suppose you have 8 lists (L1,l2,..., l8) of length n, and then calculate the worst-case time complexity:221 Group merge divided into three stages, the first phase is 8 chain list of length n, the number of operations is 2n*4The second phase is a 4-length, 2N linked list, with the number of operations being 4n*2The third stage is a 2-length linked list of 4N, with the number of operations being 8n*1so the total number of operations is n*8*3It is possible to guess half of the conclusion is that the time complexity of K's list with length N is N*k*logk
Leetcode submission results are as follows:

The procedure is as follows:

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/** * 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* head= new ListNode(0);        ListNode* p=head;//head->next为返回的指针        while(1)        {            if(l1 && l2){                if(l1->val<l2->val){                    p->next=l1;                    p=l1;                    l1=l1->next;                }                else{                    p->next=l2;                    p=l2;                    l2=l2->next;                }            }            else if(l1 && l2==NULL){                p->next=l1;                break;            }            else if(l1==NULL && l2){                p->next=l2;                break;            }            else{                break;            }        }        return head->next;    }        ListNode* mergeKLists(vector<ListNode*>& lists) {        if(lists.size()==0) return NULL;        vector<ListNode*> old_lists;//合并前的lists        vector<ListNode*> new_lists;//合并后的lists        old_lists.clear();new_lists.clear();        for(int i=0;i<lists.size();i++)            old_lists.push_back(lists[i]);        while(1)        {            if(old_lists.size()==1) break;//合并成一条链表就输出            int cnt=0;            for(int i=0;i<old_lists.size()/2;i++)            {                new_lists.push_back(mergeTwoLists(old_lists[cnt],old_lists[cnt+1]));                cnt+=2;            }            if(cnt<old_lists.size())//如果链表数是奇数，则合并完前面的两两组合后，还要加入最后一个链表                new_lists.push_back(old_lists[cnt]);            old_lists.clear();            for(int i=0;i<new_lists.size();i++)                old_lists.push_back(new_lists[i]);            new_lists.clear();        }        return old_lists[0];    }};

From for notes (Wiz)

Leetcode 23. Merge k Sorted Lists