LeetCode:合并K个排序链表

题目链接:https://leetcode-cn.com/problems/merge-k-sorted-lists/description/

合并 个排序链表,返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

思路: 两两合并就可 参考两个链表合并 https://blog.csdn.net/smile__dream/article/details/81559296

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) { 
        if(l2==null&&l1!=null){
            return l1;
        }
        if(l2!=null&&l1==null){
            return l2;
        }
        if(l1==null&&l2==null){
            return null;
        }
         ListNode mListNode,mheadNode;
       if(l1.val<l2.val)
       {
            mListNode=new ListNode(l1.val);
           mheadNode=mListNode;
           l1=l1.next;
       }else{
           mListNode=new ListNode(l2.val);
           l2=l2.next;
            mheadNode=mListNode;
       }
            
        while(l1!=null&&l2!=null){
            if(l1.val<l2.val){
                 ListNode listNode=new ListNode(l1.val);
                 mListNode.next=listNode;
                 mListNode=mListNode.next;
                 l1=l1.next;
                
            }else{
                 ListNode listNode=new ListNode(l2.val);
                 mListNode.next=listNode;
                 mListNode=mListNode.next;
                 l2=l2.next;
               
            }
        }
       while(l1!=null){
             ListNode listNode=new ListNode(l1.val);
             mListNode.next=listNode;
             mListNode=mListNode.next;
            l1=l1.next;
        }
         while(l2!=null){
             ListNode listNode=new ListNode(l2.val);
             mListNode.next=listNode;
             mListNode=mListNode.next;
            l2=l2.next;
        }
       return mheadNode;
}

    public ListNode mergeKLists(ListNode[] lists) {
        if(lists==null||lists.length==0){
            return null;
        }
      // ListNode mheadNode;
        int n=lists.length;
        while(n>1){
            int k=(n+1)/2;
            for(int i=0;i<n/2;i++)
            {
           lists[i]=mergeTwoLists(lists[i],lists[i+k]);
            }
            n=k;
        }
       return lists[0];
    }
}

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转载自blog.csdn.net/smile__dream/article/details/81560201
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