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
        if(lists.empty()){
            return nullptr;
        }
        ListNode* head = lists[0];
        for (int i = 1; i < lists.size(); i++) {
            head = Merge(head, lists[i]);
        }
        return head;
    }
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here
        if (pHead1 == nullptr) {
            return pHead2;
        }
        if (pHead2 == nullptr) {
            return pHead1;
        }
        ListNode* head;
        if (pHead1->val < pHead2->val) {
            head = pHead1;
            pHead1 = pHead1->next;
        } else {
            head = pHead2;
            pHead2 = pHead2->next;
        }
        ListNode* p = head;
        while (pHead1 != nullptr && pHead2 != nullptr) {
            if (pHead1->val < pHead2->val) {
                p->next = pHead1;
                p = p->next;
                pHead1 = pHead1->next;
            } else {
                p->next = pHead2;
                p = p->next;
                pHead2 = pHead2->next;
            }
        }
        if (pHead1 == nullptr && pHead2 != nullptr) {
            p->next = pHead2;
        } else if (pHead1 != nullptr && pHead2 == nullptr) {
            p->next = pHead1;
        } else {
            p->next = nullptr;
        }
        return head;
    }
};