蜗牛慢慢爬 LeetCode 23. Merge k Sorted Lists [Difficulty: Hard]

题目

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

翻译

合并k个有序的链表

Hints

Related Topics: LinkedList, Divide and Conquer, Heap

Solution1：Divide and Conquer

参考归并排序-维基百科

思路：分治，先分成两个子任务，然后递归子任务，最后回溯回来..这里就可以将k个链表分成两半，再继续划分，直到最后还剩下2个链表就利用之前做的 Merge Two Sorted Lists 来合并

Solution2：Heap

用到了堆的数据结构（太菜了想不到，还是要学习一个）

学习一下堆的数据结构，维护一个大小为k的堆，每次取堆顶的最小元素放到结果中，然后再把该元素的下一个元素放到堆中，重新维护好；因为每个链表都是有序的，每次都是取堆中最小的元素，这样当所有的链表都读完的时候，result链表中就是所有的元素从小到大排列了

Solution2要读一遍所有元素：即 k*n 次，读取元素又将新元素插入堆中是 logn的复杂度，所有最后时间复杂度是O(nk logn)..

代码

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
//Solution1
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return Divide(lists, 0, lists.length-1);
    }    public static ListNode Divide(ListNode[] lists, int start, int end){
        if(start==end)  return lists[start];
        if(start<end){
            int middle = (start+end)/2;
            ListNode l1 = Divide(lists,start,middle);
            ListNode l2 = Divide(lists,middle+1,end);
            return merge2Lists(l1, l2);
        }else
            return null;
    }    public static ListNode merge2Lists(ListNode l1,ListNode l2){
        if(l1==null)    return l2;
        if(l2==null)    return l1;        if(l1.val<l2.val){
            l1.next = merge2Lists(l1.next,l2);
            return l1;
        }else{
            l2.next = merge2Lists(l1,l2.next);
            return l2;
        }
    }
}

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
//Solution2
from Queue import PriorityQueue
class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        dummy = curr = ListNode(None)
        q = PriorityQueue()        for node in lists:
            if node:    q.put((node.val,node));
        while q.qsize()>0:
            curr.next = q.get()[1]
            curr = curr.next
            if curr.next:
                q.put((curr.next.val,curr.next))
        return dummy.next