【Lintcode】104.Merge k Sorted Lists

题目：

Merge k sorted linked lists and return it as one sorted list.

Analyze and describe its complexity.

Example

Given lists:

[
  2->4->null,
  null,
  -1->null
],

return -1->2->4->null.

题解：

Solution 1 ()

class Solution {
public:
    struct compare {
        bool operator() (const ListNode* a, const ListNode* b) {
            return a->val > b->val;
        }
    };
    ListNode* mergeKLists(vector<ListNode *> &lists) {
        priority_queue<ListNode*, vector<ListNode*>, compare> q;
        for (auto l : lists) {
            if (l) {
                q.push(l);
            }
        }
        ListNode* head = nullptr, *pre = nullptr, *tmp = nullptr;
        while (!q.empty()) {
            tmp = q.top();
            q.pop();
            if(!pre) {
                head = tmp;
            } else {
                pre->next = tmp;
            }
            pre = tmp;
            if (tmp->next) {
                q.push(tmp->next);
            }
        }
        
        return head;
    }
};

 

Solution 2 ()

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode *> &lists) {
        if (lists.empty()) {
            return nullptr;
        }
        int n = lists.size();
        while (n > 1) {
            int k = (n + 1) / 2;
            for (int i = 0; i < n / 2; ++i) {
                lists[i] = mergeTwoLists(lists[i], lists[i + k]);
            }
            n = k;
        }
        return lists[0];
    }
    
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1);
        ListNode* cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if (l1) {
            cur->next = l1;
        } else {
            cur->next = l2;
        }
        
        return dummy->next;
    }
};

 

Solution 3 ()

class Solution {
public:
    static bool heapComp(ListNode* a, ListNode* b) {
        return a->val > b->val;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) { //make_heap
        ListNode head(0);
        ListNode *curNode = &head;
        vector<ListNode*> v;   
        for(int i =0; i<lists.size(); i++){
            if(lists[i]) v.push_back(lists[i]);
        }
        make_heap(v.begin(), v.end(), heapComp); //vector -> heap data strcture

        while(v.size()>0){
            curNode->next=v.front();
            pop_heap(v.begin(), v.end(), heapComp); 
            v.pop_back(); 
            curNode = curNode->next;
            if(curNode->next) {
                v.push_back(curNode->next); 
                push_heap(v.begin(), v.end(), heapComp);
            }
        }
        return head.next;
    }
};