023. Merge k Sorted Lists

 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 };