[LeetCode] Merge k Sorted Lists

 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

归并K个有序链表

思路1：使用multiset存储每一个出现的元素。再将这些元素连接起来即可（vector也可以）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        multiset<int> mset;
        for (auto& list : lists)
        {
            ListNode* cur = list;
            while (cur)
            {
                mset.insert(cur->val);
                cur = cur->next;
            }
        }
        ListNode* head = nullptr;
        ListNode* node = nullptr;
        for (auto it = mset.begin(); it != mset.end(); ++it)
        {
            if (node)
            {
                node->next = new ListNode(*it);
                node = node->next;
            }
            else
            {
                node = new ListNode(*it);
                head = node;
            }
        }
        return head;
    }
};

思路2：堆排序，使用优先队列

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        auto cmp = [](const ListNode* a, const ListNode* b) { return a->val > b->val; };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp)> min_heap(cmp);
        for (auto& list : lists)
            if (list)
                min_heap.push(list);
        ListNode* dummy = new ListNode(-1);
        ListNode* cur = dummy;
        while (!min_heap.empty())
        {
            ListNode* node = min_heap.top();
            min_heap.pop();
            cur->next = node;
            cur = cur->next;
            if (node->next)
                min_heap.push(node->next);
        }
        return dummy->next;
    }
};

思路三：利用分治+递归

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty())
            return NULL;
        while (lists.size() > 1)
        {
            lists.push_back(mergeTwoLists(lists[0], lists[1]));
            lists.erase(lists.begin());
            lists.erase(lists.begin());
        }
        return lists[0];
    }
    
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
    {
        if (l1 == NULL)
            return l2;
        if (l2 == NULL)
            return l1;
        if (l1->val < l2->val)
        {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
        else
        {
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};