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) {}
 * };
 */
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* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        ListNode *head = new ListNode(0);

        if(lists.empty())
        {
            return nullptr;
        }

        if(lists.size() == 1)
        {
            return lists[0];
        }
        
       for(int i=0;i<lists.size()-1;i++)
        {
            ListNode *pHead1 = lists[i];
            ListNode *pHead2 = lists[i+1];  
            ListNode *nHead = head;
            while(pHead1 && pHead2)
            {
                if(pHead1->val < pHead2->val)
                {
                    nHead->next = pHead1;
                    pHead1 = pHead1->next;
                }
                else 
                {
                    nHead->next = pHead2;
                    pHead2 = pHead2->next;
                }
                nHead = nHead->next;
            }
            nHead->next = (pHead1?pHead1:pHead2);
            lists[i+1] = head->next;
        }

        return head->next;

    }
};