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