leetcode 23. 合并K个排序链表 JAVA

题目:

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

解题思路:

使用归并将链表两两合并

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length == 1 )
            return lists[0];
        else if(lists == null)
            return null;
        return MSort(lists,0,lists.length - 1);
    }
    
    public ListNode MSort(ListNode[] lists, int low, int high){ 
        if(low < high){  
            int mid = (low+high)/2;
            ListNode leftlist = MSort(lists,low,mid);
            ListNode rightlist = MSort(lists,mid+1,high);
            return mergeTwoLists(leftlist,rightlist);  
        }  
        else if(low == high)
        {
            return lists[low];//
        }
        else 
            return null; //
    }
    
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode cur = head;
        ListNode p = l1, q = l2;
        while(p != null && q != null)
        {
            if(p.val <= q.val)
            {
                cur.next = new ListNode(p.val);
                cur = cur.next;
                p = p.next;
            }
            else
            {
                cur.next = new ListNode(q.val);
                cur = cur.next;
                q = q.next;
            }             
        }
        while(p != null)
        {
            cur.next = new ListNode(p.val);
            cur = cur.next;
            p = p.next;
        }
        while(q != null)
        {
            cur.next = new ListNode(q.val);
            cur = cur.next;
            q = q.next;
        }
        return head.next;
    }
    
}
原文地址:https://www.cnblogs.com/yanhowever/p/10457544.html