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* mergeNode(ListNode* node1, ListNode* node2) {
        ListNode* new_node = new ListNode(0);
        ListNode* new_node_cur = new_node;
        while (node1&&node2) {
            if(node1->val<node2->val){
                new_node_cur->next = new ListNode(node1->val);
                node1 = node1->next;
            } else {
                new_node_cur->next = new ListNode(node2->val);
                node2 = node2->next;
            }
            new_node_cur = new_node_cur->next;
        }
        new_node_cur->next = node1?node1:node2;
        return new_node->next;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int size = lists.size();
        if (size == 0) return nullptr;
        ListNode* node = lists[0];
        for(int i=1;i<size;++i){
            ListNode* temp_node = lists[i];
            node = mergeNode(node, temp_node);
        }
        return node;
    }
};