LeetCode--023--合并K个排序链表

合并 k 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

方法1：分治

time:O(nlogk) where k is the number of linked list

space:O(n)

 /**
  * 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 == null || lists.length == 0)return null;
         return sort(lists,0,lists.length - 1);
     }
     public ListNode sort(ListNode[] lists,int lo,int hi){
         if(lo >= hi) return lists[lo];//奇数个的时候，两两合并会多出一个，此语句用于返回该list,在下一次进行合并
         int mid = (hi - lo) / 2 + lo;
         ListNode l1 = sort(lists,lo,mid);
         ListNode l2 = sort(lists,mid+1,hi);
         return merge(l1,l2);
     }
     public ListNode merge(ListNode l1,ListNode l2){
         if(l1 == null) return l2;
         if(l2 == null) return l1;
         if(l1.val < l2.val){
             l1.next = merge(l1.next,l2);
             return l1;
         }
         l2.next = merge(l1,l2.next);
         return l2;
     }
 }

方法2：优先级队列

 /**
  * 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 == null || lists.length == 0)return null;
         PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length,(a,b)->a.val-b.val);
         ListNode dummy = new ListNode(0);
         ListNode cur = dummy;

         for(ListNode list : lists){
             if(list != null){
                 queue.add(list);
             }
         }
         while(!queue.isEmpty()){
             cur.next = queue.poll();
             cur = cur.next;
             if(cur.next != null){
                 queue.add(cur.next);
             }
         }
         return dummy.next;
     }

 }

2019-04-19 10:07:41