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* mergeList(ListNode *A, ListNode *B){
        ListNode *res = new ListNode(0);
        ListNode *p = res;
        while(A != nullptr && B != nullptr){
            if( A->val < B->val ){
                p->next = A;
                A = A->next;
            }else{
                p->next = B;
                B = B->next;
            }
            p = p->next;
        }
        if (A != nullptr){
            p->next = A;
        }
        if (B != nullptr) {
            p->next = B;
        }
        return res->next;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        int lists_num = lists.size();
        if(lists_num == 0){
            return nullptr;
        }
        vector<vector<int>> vals_array;

        ListNode *A = lists[0];
        ListNode *res = lists[0];
        for (int i = 1; i < lists_num; i++)
        {
            res = mergeList(A,lists[i]);
            A = res;            
        }
        return res;
    }

};