leetcode23. Merge k Sorted Lists

Posted 2020-08-03 godlei

题目描述：

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

解题分析：

解题思路很简单，就是先两两排序，不断减小链表数量，最后将所有数组织成一个有序链表

具体代码：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10    public static ListNode mergeKLists(ListNode[] lists) {
11         if(lists.length==0)
12             return null;
13         if(lists.length==1)
14             return lists[0];
15         int from=0;
16         int to=lists.length-1;
17         //两两排序时使用类似二分查找的方法缩小时间复杂度
18         while(from!=to){
19             //from+to等于偶数
20             if(((from+to)&1)==0){
21                 int mid=(from+to)>>1;
22                 for(int i=0;i<mid;i++){
23                     lists[i]=mergeTwoLists(lists[i], lists[mid+1+i]);
24                 }
25                 to=mid;
26             }
27             else{
28                 int mid=(from+to)>>1;
29                 for(int i=0;i<=mid;i++){
30                     lists[i]=mergeTwoLists(lists[i], lists[mid+1+i]);
31                 }
32                 to=mid;
33             }
34         }
35         return lists[0];
36     }
37     //和并两个链表的方法
38     public static ListNode mergeTwoLists(ListNode head1, ListNode head2) {
39         if(head1==null)
40             return head2;
41         if(head2==null)
42             return head1;
43         ListNode head=null;
44         ListNode current=null;
45         if(head1.val<=head2.val){
46             head=head1;
47             head1=head1.next;
48             head.next=null;
49         }
50         else{
51             head=head2;
52             head2=head2.next;
53             head.next=null;
54         }
55         current=head;
56         while(head1!=null&&head2!=null){
57             if(head1.val<=head2.val){
58                 current.next=head1;
59                 current=current.next;
60                 head1=head1.next;
61                 current.next=null;
62             }
63             else{
64                 current.next=head2;
65                 current=current.next;
66                 head2=head2.next;
67                 current.next=null;
68             }
69         }
70         if(head1!=null){
71             current.next=head1;
72         }
73         if(head2!=null){
74             current.next=head2;
75         }
76         return head;
77     }
78 }