【leetcode】23. Merge k Sorted Lists

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：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
   public static ListNode mergeKLists(ListNode[] lists) {
        if(lists.length==0)
            return null;
        if(lists.length==1)
            return lists[0];
        int from=0;
        int to=lists.length-1;
        //两两排序时使用类似二分查找的方法缩小时间复杂度
        while(from!=to){
            //from+to等于偶数
            if(((from+to)&1)==0){
                int mid=(from+to)>>1;
                for(int i=0;i<mid;i++){
                    lists[i]=mergeTwoLists(lists[i], lists[mid+1+i]);
                }
                to=mid;
            }
            else{
                int mid=(from+to)>>1;
                for(int i=0;i<=mid;i++){
                    lists[i]=mergeTwoLists(lists[i], lists[mid+1+i]);
                }
                to=mid;
            }
        }
        return lists[0];
    }
    //和并两个链表的方法
    public static ListNode mergeTwoLists(ListNode head1, ListNode head2) {
        if(head1==null)
            return head2;
        if(head2==null)
            return head1;
        ListNode head=null;
        ListNode current=null;
        if(head1.val<=head2.val){
            head=head1;
            head1=head1.next;
            head.next=null;
        }
        else{
            head=head2;
            head2=head2.next;
            head.next=null;
        }
        current=head;
        while(head1!=null&&head2!=null){
            if(head1.val<=head2.val){
                current.next=head1;
                current=current.next;
                head1=head1.next;
                current.next=null;
            }
            else{
                current.next=head2;
                current=current.next;
                head2=head2.next;
                current.next=null;
            }
        }
        if(head1!=null){
            current.next=head1;
        }
        if(head2!=null){
            current.next=head2;
        }
        return head;
    }
}