技術標籤：刷題

難度：中等。
重點在連結串列的操作，不要搞混了。

正確解法：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
 

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int n = lists.size();
        if(n == 0)return NULL;
        if(n == 1)return lists[0];

        int is_empty = 0;
        vector<ListNode*> temp = lists;
        int flag = 1;
        ListNode* list,
 
 *result;
        int it = 0;
        while(is_empty < n){
            int now_list = 0;
            int now_num = 100001;
            for(int i = 0; i < n; i++){
                if(temp[i]){
                    int k = temp[i]->val;
                    if(now_num > k){
                        now_num =
 
 k;
                        now_list = i;
                    }
                }
                else if(it == 0){
                    is_empty++;
                }
            }
            if(is_empty == n)return NULL;
            if(flag){
                list = temp[now_list];
                result = temp[now_list];
                flag = 0;
            }
            else{
                list->next = temp[now_list];
                list = list->next;
            }
            if(!temp[now_list]->next){
                is_empty++;
            }
            temp[now_list] = temp[now_list]->next;
            if(it == 0)it = 1;
        }
        return result;
    }
};