LeetCode——023

/*
23. Merge k Sorted Lists My Submissions QuestionEditorial Solution
Total Accepted: 82416 Total Submissions: 354043 Difficulty: Hard
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

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*/
/*
解题思路：
采用二路归并的方法
先两两结合进行合并把结果放到前一个链表指针的位置。一趟之后，指针数组会减少一半成为n/2,然后再对这n/2链表进行合并，依次这样下去，知道最后链表指针个数为1即为我们所要的结果。

注意：在两个链表合并的时候，若要合并的是同一个链表，则直接返回不需要做合并。
*/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    //合并两个已经排好序的链表
    ListNode* mergeTwoLists(ListNode* l1,ListNode* l2){

        if(l1==l2)return l1;
        if(l1==NULL)return l2;
        if(l2==NULL)return l1;
        ListNode* dummy=new ListNode(-1);
        ListNode* p=dummy;
        while(l1&&l2){

            if(l1->val<l2->val){
                p->next=l1;
                l1=l1->next;
            }
            else{
                p->next=l2;
                l2=l2->next;
            }
            p=p->next;
        }

        if(l1)p->next=l1;
        if(l2)p->next=l2;
        return dummy->next;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {

        if(lists.empty())return NULL;
       while(lists.size()>1){
           int left=0,right=lists.size()-1;
           while(left<=right){
               lists[left]=mergeTwoLists(lists[left],lists[right]);
               ++left;
               --right;
           }
           lists.resize(left);
       }
       return lists[0];
    }
};