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/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {
   public:    ListNode* mergetwo(ListNode* l1, ListNode *l2){
           ListNode *out = new ListNode(0), *L;        L=out;        while(l1!=nullptr&&l2!=nullptr){
               if(l1->val
   
    val){
                   L->next=l1;                L=l1;                l1=l1->next;            }else{
                   L->next =l2;                L=l2;                l2=l2->next;            }        }        L->next = (l1?l1:l2);        return out->next;    }    ListNode* merge(vector
    
     & lists, int l, int n)    {
           if(l==n) return lists[l];        if(l>n) return nullptr;        int mid = (l+n)>>1;        return mergetwo(merge(lists, l, mid), merge(lists, mid+1, n));    }    ListNode* mergeKLists(vector
     
      & lists) {
           return merge(lists, 0, lists.size()-1);    }};
     
    
   

关键词：两两循环排序

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {
   public:    ListNode* mergetwo(ListNode* l1, ListNode *l2){
           ListNode *out = new ListNode(0), *L;        L=out;        while(l1!=nullptr&&l2!=nullptr){
               if(l1->val
   
    val){
                   L->next=l1;                L=l1;                l1=l1->next;            }else{
                   L->next =l2;                L=l2;                l2=l2->next;            }        }        L->next = (l1?l1:l2);        return out->next;    }    ListNode* mergeKLists(vector
    
     & lists) {
           ListNode *out = nullptr;        for(int i=0;i
    
   

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