题解 | 合并k个已排序的链表

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
 #include <vector>
// 声明std命名空间，避免写std::vector（也可以不写，把vector改成std::vector）
using namespace std;
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param lists ListNode类vector 
     * @return ListNode类
     */
         ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
 
        if(pHead1==nullptr and pHead2==nullptr)
        return pHead1;
        else if(pHead1==nullptr and pHead2!=nullptr)
        return pHead2;
        else if(pHead2==nullptr and pHead1!=nullptr)
        return pHead1;
        else{
            ListNode* Head;
            ListNode* curr1=pHead1;
            ListNode* curr2=pHead2;
            if(pHead1->val<pHead2->val){
            Head=pHead1;
            curr1=curr1->next;
            }
            else{
               Head=pHead2;
               curr2=curr2->next;
            }
            ListNode* curr=Head;
            while(curr1!=nullptr and curr2!=nullptr){
                if(curr1->val<curr2->val){
                curr->next=curr1;
                curr=curr1;
                curr1=curr1->next;
                }
                else{
                    curr->next=curr2;
                    curr=curr2;
                    curr2=curr2->next;
                }
            }
            if(curr1==nullptr)
            curr->next=curr2;
            else if(curr2==nullptr)
            curr->next=curr1;
            return Head;
        }
        }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int k=lists.size();
        if(k==0)
        return nullptr;
        for(int i=0;i<k-1;i++){
            lists[0]=Merge(lists[0],lists[i+1]);
        }
        return lists[0];
    }
         };