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* Merge(ListNode* pHead1, ListNode* pHead2) {
        if (pHead1 == nullptr)return pHead2;
        if (pHead2 == nullptr) return pHead1;
        ListNode* newnode;
        if (pHead1->val <= pHead2->val) {
            newnode = pHead1;
            pHead1 = pHead1->next;
        } else {
            newnode = pHead2;
            pHead2 = pHead2->next;
        }
        newnode->next = Merge(pHead1, pHead2);
        return newnode;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode* res = nullptr;
        for (auto * list : lists) {
            res = Merge(res, list);
        }
        return res;

    }
};