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  • 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;
        }
        
    }
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  • 原文地址:https://www.cnblogs.com/yanhowever/p/10457544.html
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