使用大根堆里存储链表节点值的相反数——

priority_queue[HTML_REMOVED]> heap;

class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        priority_queue<pair<int, ListNode*>> heap;
        auto dummy = new ListNode(-1), tail = dummy;

        for (auto list : lists) {
            if (list) heap.push({-list->val, list});
        }

        while (heap.size()) {
           auto t = heap.top();
           heap.pop();
           tail = tail->next = t.second;
           auto p = t.second->next;
           if (p) heap.push({-p->val, p});
        }
        return dummy->next;
    }
};

二分治合并（基于二路归并思想） 参考题解1

参考题解2 使用默认大根堆，但是需要存储值的相反数
在堆中存储的是一对pair，其中第一个值就是指针指向的值，第二个值为当前指针。当每个元素在小根堆中排序时，默认根据第一维，即对每个指针指向结点的值进行排序。

参考题解3 我不是张小白 CSDN上的题解

class Solution {
public:

    ListNode* merge2Lists(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1), tail = dummy;
        while (l1 && l2) {
            if (l1->val > l2->val) {
                tail = tail->next = l2;
                l2 = l2->next;
            } 
            else {
                tail = tail->next = l1;
                l1 = l1->next;
            }
        }
        tail = tail->next = (l1? l1 : l2);
        return dummy->next;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.size() == 0) return NULL;
        if (lists.size() == 1) return lists[0];
        int mid = lists.size() / 2;
        vector<ListNode*> left = vector<ListNode*>(lists.begin(), lists.begin() + mid);
        vector<ListNode*> right = vector<ListNode*>(lists.begin() + mid, lists.end());
        ListNode* l1 = mergeKLists(left);
        ListNode* l2 = mergeKLists(right);
        return merge2Lists(l1, l2);
    }
};

yxc代码参考

class Solution {
public:

    struct Cmp {
        bool operator() (ListNode* a, ListNode* b) {
            return a->val > b->val;
        }
    };

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, Cmp> heap;
        auto dummy = new ListNode(-1), tail = dummy;
        for (auto l : lists) if (l) heap.push(l);

        while (heap.size()) {
            auto t = heap.top();
            heap.pop();

            tail = tail->next = t;
            if (t->next) heap.push(t->next);
        }

        return dummy->next;
    }
};

小根堆的另一种定义方式

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
                auto cmp = [](ListNode*& a, ListNode*& b) {
            return a->val > b->val;
        };
        priority_queue<ListNode*, vector<ListNode*>, decltype(cmp) > q(cmp);
        for (auto node : lists) {
            if (node) q.push(node);
        }
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (!q.empty()) {
            auto t = q.top(); q.pop();
            cur->next = t;
            cur = cur->next;
            if (cur->next) q.push(cur->next);
        }
        return dummy->next;
    }
};

另一种归并方式写法
先将K个链表分成两个部分，先合并两个部分，在合并这两个部分返回的链表。

ListNode* mergeKLists(vector<ListNode*>& lists) {
        int n = lists.size();
        if(n == 0) return NULL;
        return merge(lists,0,n  - 1);

    }
    ListNode* merge(vector<ListNode*>& lists,int begin,int end)
    {
        if(begin == end)
            return lists[begin];
        else{
            int mid = (begin + end) /2;
            ListNode* l1 = merge(lists,begin,mid);
            ListNode* l2 = merge(lists,mid+1,end);
            ListNode* dummy = new ListNode(0);
            ListNode* cur = dummy;
            while(l1 && l2)
            {
                if(l1->val < l2->val)
                {
                    cur->next = new ListNode(l1->val);
                    l1 = l1->next;
                }else{
                    cur->next = new ListNode(l2->val);
                    l2 = l2->next;
                }
                cur = cur->next;
            }
            cur->next = (l1)?l1:l2;
            return dummy->next;
        }
    }