链表归并排序

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
         if(pHead1==nullptr)
         {
             return pHead2;
         }
         if(pHead2==nullptr)
         {
             return pHead1;
         }        
        ListNode* res=nullptr;
        if(pHead1->val>=pHead2->val)
        {
            res=pHead2;
            res->next=Merge(pHead1,pHead2->next);
        }
        else
        {
            res=pHead1;
            res->next=Merge(pHead1->next,pHead2);
        }
        return res;
    }
};

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

/**
 * 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  dummy(-1); 
        ListNode *result = &dummy;
        while(l1 && l2)
        {
            if(l1->val <= l2->val)
            {
                result->next = l1;
                result = result->next;
                l1 = l1->next;
            }
            else
            {
                result->next = l2;                
                result = result->next;
                l2 = l2->next;
            }            
        }
        if(l1)
            result->next = l1;
        else if(l2)
            result->next = l2;
        return dummy.next;
    }

    ListNode *mergeKLists(vector<ListNode *> &lists) 
    {
        if (lists.empty()) return NULL;
        int len = lists.size();
        while (len > 1) 
        {
            for (int i = 0; i < len / 2; ++i) 
            {
                lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
            }
            len = (len + 1) / 2;
        }
        return lists.front();
    }
};