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23. Merge k Sorted Lists

▶ 合并 k 个已经排好序的单链表

● 代码，28 ms，二路归并递归版，时间复杂度 O(k n log k)

 1 class Solution
 2 {
 3 public:
 4     ListNode *mergeTwoLists(ListNode* l1, ListNode* l2)
 5     {
 6         if (l1 == nullptr)
 7             return l2;
 8         if (l2==nullptr)
 9             return l1;
10         if (l1->val <= l2->val)
11         {
12             l1->next = mergeTwoLists(l1->next, l2);
13             return l1;
14         }
15         else
16         {
17             l2->next = mergeTwoLists(l1, l2->next);
18             return l2;
19         }
20     }
21     ListNode *mergeKLists(vector<ListNode *> &lists)
22     {
23         if (lists.empty())
24             return nullptr;
25         int i, len;
26         for (len = lists.size(); len > 1; len = (len + 1) / 2)
27         {
28             for (i = 0; i < len / 2; i++)
29                 lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
30         }
31         return lists.front();
32     }
33 };

● 代码，30 ms，二路归并非递归版，时间复杂度 O(k n log k)

 1 class Solution
 2 {
 3 public:
 4     ListNode *mergeTwoLists(ListNode* l1, ListNode* l2)
 5     {
 6         ListNode  dummy(-1), *result = &dummy;
 7         for(;l1!=nullptr &&l2!=nullptr;)
 8         {
 9             if (l1->val <= l2->val)
10             {
11                 result->next = l1;
12                 l1 = l1->next;
13             }
14             else
15             {
16                 result->next = l2;
17                 l2 = l2->next;
18             }
19             result = result->next;
20         }
21         if (l1)
22             result->next = l1;
23         else if (l2)
24             result->next = l2;
25         return dummy.next;
26     }
27     ListNode *mergeKLists(vector<ListNode *> &lists)
28     {
29         if (lists.empty())
30             return NULL;
31         int i, len;
32         for (len = lists.size(); len > 1; len = (len + 1) / 2)
33         {
34             for (i = 0; i < len / 2; i++)
35                 lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);            
36         }
37         return lists.front();
38     }
39 };

● 代码，29 ms，优先队列，最快的解法算法与之相同，时间复杂度 O(N log k)

 1 class Solution
 2 {
 3 public:
 4     struct compare
 5     {
 6         bool operator()(const ListNode* l, const ListNode* r)
 7         {
 8             return l->val > r->val;
 9         }
10     };
11     ListNode *mergeKLists(vector<ListNode *> &lists)
12     {
13         priority_queue<ListNode *, vector<ListNode *>, compare> q;
14         ListNode *result, *tail;
15         for (auto l : lists) // 把各单链表首元素加入优先队中
16         {
17             if (l)
18                 q.push(l);
19         }
20         if (q.empty())
21             return NULL;
22         result = q.top(), q.pop();  // 取出头部
23         if (result->next)           
24             q.push(result->next);
25         for (tail = result; !q.empty();)
26         {
27             tail->next = q.top(), q.pop();// 每次取出一个元素，若取出的元素有后继，则将后继加入优先队中
28             tail = tail->next;
29             if (tail->next)
30                 q.push(tail->next);
31         }
32         return result;
33     }
34 };

● 代码，29 ms，使用堆，时间复杂度 O(N log k)

 1 class Solution
 2 {
 3 public:
 4     static bool heapComp(ListNode* a, ListNode* b)
 5     {
 6         return a->val > b->val;
 7     }
 8     ListNode* mergeKLists(vector<ListNode*>& lists)
 9     {
10         ListNode head(0);
11         ListNode *curNode = &head;
12         vector<ListNode*> v;
13         for (int i = 0; i<lists.size(); i++)
14         {
15             if (lists[i])
16                 v.push_back(lists[i]);
17         }
18         make_heap(v.begin(), v.end(), heapComp);// 建堆
19         for (;v.size()>0;)
20         {
21             curNode->next = v.front();
22             pop_heap(v.begin(), v.end(), heapComp);
23             v.pop_back();
24             curNode = curNode->next;
25             if (curNode->next)
26             {
27                 v.push_back(curNode->next);
28                 push_heap(v.begin(), v.end(), heapComp);
29             }
30         }
31         return head.next;
32     }
33 };

 

posted on 2018-01-29 23:11  爨爨爨好  阅读(155)  评论(0)  编辑  收藏  举报