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:

    ListNode* merge(ListNode *node1, ListNode *node2) {
        ListNode head(0);
        ListNode* cur = &head;
        while (node1 && node2) {
            if (node1->val < node2->val) {
                cur->next = node1;
                node1 = node1->next;
            } else {
                cur->next = node2;
                node2 = node2->next;
            }
            cur = cur->next;
        }
        if (node1) {
            cur->next = node1;
        }
        if (node2) {
            cur->next = node2;
        }
        return head.next;
    }

    /**
     * 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) {
        ListNode* cur = nullptr;
        for (int i = 0; i < lists.size(); i++) {
            cur = merge(cur, lists[i]);
        }
        return cur;
    }
};