1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* mergeKLists(vector<ListNode*>& lists) {
　　　　　　 // 注意这里的判空情况[[]]12         // 寻找最小值节点13         ListNode* ptr = nullptr;14         ListNode* pHead = ptr;15         if (lists.size() == 0) return pHead;16         else {17             int minIndex = -1;18             for (int i = 0; i < lists.size(); ++i) {19                 if (lists[i] != nullptr) {20                     if (minIndex == -1) minIndex = i;21                     else {22                         if (lists[i]->val < lists[minIndex]->val) minIndex = i;23                     }24                 }25             }26             if (minIndex != -1) {27                 ptr = lists[minIndex];28                 lists[minIndex] = lists[minIndex]->next;29             }30             //cout << ptr->val << endl;31             pHead = ptr; // 头指针32             bool flag = false; // 判断是否所有的点都已经排序33             while (!flag) {34                 flag = true;35                 int index = -1; // 记录最小节点下标36                 for (int i = 0; i < lists.size(); ++i) {37                     if (lists[i] != nullptr) {38                         flag = false;39                         if (index == -1) index = i;40                         else {41                             if (lists[i]->val < lists[index]->val) index = i;42                         }43                     }44                 }45                 46                 if (index != -1) {47                     //cout << lists[index]->val << endl;48                     ptr->next = lists[index];49                     ptr = ptr->next;50                     lists[index] = lists[index]->next;51                 }52             }53             return pHead;54         }55     }56 };