【Hard】23. Merge k Sorted Lists

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

中心思想：（此题应有多解）

之前做过merge sorted lists的题，用那一题的方法，写一个Helper function，然后将此列表中的数组一一merge。此方法time complexity可能不够好。

/** * 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) {        int n = lists.size();        ListNode* result = NULL;        if (n ==0){            return result;        }else if (n == 1){            return lists[0];        }                        for (int i = 0; i < n; i++){            result = merge(result, lists[i]);        }                return result;                        }            ListNode* merge(ListNode* l1, ListNode* l2){        if (l1 == NULL)            return l2;        else if (l2 == NULL)            return l1;                   ListNode* head=NULL;                if(l1->val <= l2->val){            head = l1;            l1 = l1->next;        }else{            head = l2;            l2 = l2->next;                    }                ListNode* ret=head;                while(l1 != NULL && l2 != NULL){            if (l1->val <= l2->val){                ret->next = l1;                l1 = l1->next;            }            else if (l1->val > l2->val){              ret->next = l2;                l2 = l2->next;            }            ret = ret->next;        }                if (l1 == NULL)            ret->next = l2;        else if (l2 == NULL)            ret->next = l1;                    return head;    }};

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