23. 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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 思路：类似归并排序，每个链表已经排好序了，现在只需要将各个链表合并成一个链表。要点：分而治之，最后合并。

  1. /**
  2.  * Definition for singly-linked list.
  3.  * struct ListNode {
  4.  *     int val;
  5.  *     ListNode *next;
  6.  *     ListNode(int x) : val(x), next(NULL) {}
  7.  * };
  8.  */
  9. class Solution {
  10. private:
  11.     ListNode * merge_help(vector<ListNode*>&lists,int l,int r){
  12.         if(l>=r)
  13.             return lists[l];
  14.         int mid=(l+r)/2;
  15.         ListNode * left =merge_help(lists,l,mid);
  16.         ListNode * right=merge_help(lists,mid+1,r);
  17.         ListNode * temp =new ListNode (-1);
  18.         ListNode *cur =temp;
  19.         while(left&&right){
  20.             if(left->val<=right->val){
  21.                 cur->next =left;
  22.                 left=left->next;
  23.             }
  24.             else{
  25.                 cur->next=right;
  26.                 right=right->next;
  27.             }
  28.             cur=cur->next;
  29.         }
  30.         if(left)
  31.             cur->next=left;
  32.         if(right)
  33.             cur->next=right;
  34.         return temp->next;
  35.         
  36.     }
  37. public:
  38.     ListNode* mergeKLists(vector<ListNode*>& lists) {
  39.         if(lists.size()==0)
  40.             return NULL;
  41.         return merge_help(lists,0,lists.size()-1);
  42.             
  43.     }
  44.     
  45.     
  46. };





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原文地址：https://www.cnblogs.com/zhoudayang/p/5293434.html