Merge k Sorted Lists

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

 

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这道题试了多种方法，有堆排序和分zhi链表排序

class Solution {
public:
    //这个会超时是因为排序到后面，其中一个链表会越来越长，导致合并的时间越来越长
    ListNode *_mergeKLists(vector<ListNode *> &lists) {
        if (lists.empty()) return 0;
        if (lists.size() == 1) return lists.back();

        vector<ListNode *> l;
        while (!lists.empty()) {
                if (lists.size() == 1) {
                        l.push_back(lists.back());
                        lists.pop_back();
                }
                else {
                        ListNode *p = lists.back();
                        lists.pop_back();
                        ListNode *q = lists.back();
                        lists.pop_back();

                        l.push_back(mergeTwoLists(p, q));
                }
        }

        return mergeKLists(l);
    }
    
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.empty()) {
            return NULL;
        }

        while (!lists.empty()) {
            if (1 == lists.size()) {
                return lists[0];
            }
            ListNode *p1 = lists.back();
            lists.pop_back();
            ListNode *p2 = lists.back();
            lists.pop_back();
            lists.push_back(mergeTwoLists(p1, p2));
        }
    }
　　
    ListNode* mergeTwoLists(struct ListNode *l1, struct ListNode *l2){
        if (l1 == NULL && l2 == NULL) {
            return NULL;
        } else if (l1 == NULL) {
            return l2;
        } else if (l2 == NULL) {
            return l1;
        }

        ListNode* phead;
        ListNode* node;
        ListNode* p1 = l1;
        ListNode* p2 = l2;
        if (p1->val <= p2->val) {
            phead = p1;
            p1 = p1->next;
        } else {
            phead = p2;
            p2 = p2->next;
        }
        node = phead;

        while (p1 != NULL && p2 != NULL) {
            if (p1->val <= p2->val) {
                node->next = p1;
                p1 = p1->next;
            } else {
                node->next = p2;
                p2 = p2->next;
            }
            node = node->next;
        }

        node->next = p1 ? p1 : p2;

        return phead;
    }

};