merge k sorted lists

来源：互联网发布：iphone视频软件编辑：程序博客网时间：2024/05/16 08:53

题目地址：https://oj.leetcode.com/problems/merge-k-sorted-lists/

这个题目其实就是merge two sorted lists 的升级系列，最直接的方法就是选取一个ListNode 与剩下的每一个ListNode调用merge two sorted lists 方法。

public class Solution {    public ListNode mergeKLists(List<ListNode> lists) {    if(lists.size() ==0)    return null;    ListNode Pmerge = lists.get(0);    for(int i=1;i<lists.size();i++){    Pmerge = mergeTwoLists(Pmerge, lists.get(i));    }    return Pmerge;    }    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {ListNode Pmerge = null;if(l1 == null){return l2;}if(l2 == null){return l1;}if(l1.val<l2.val){Pmerge = l1;Pmerge.next = mergeTwoLists(l1.next, l2);}else {Pmerge = l2;Pmerge.next = mergeTwoLists(l1, l2.next);}return Pmerge;}}

这样能过通过在线测试，但是反过来想我们会发现这效率肯定不高，如果有n 个ListNode ，每个长度为K,时间复杂度为O（n *K^2）. 很容易想到我们知道一类排序叫归并排序，两两递归到最小元素排序然后回溯回来实现整体排序，进行细微改动变换为利用归并排序。

public class Solution {    public ListNode mergeKLists(List<ListNode> lists) {    if(lists.size() ==0)    return null;    return helper(lists, 0, lists.size()-1);    }        public ListNode helper(List<ListNode> lists ,int l ,int r){    if(l<r){    int mid = (l+r)/2;    return mergeTwoLists(helper(lists, l, mid), helper(lists, mid+1, r));    }    return lists.get(l);    }    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {ListNode Pmerge = null;if(l1 == null){return l2;}if(l2 == null){return l1;}if(l1.val<l2.val){Pmerge = l1;Pmerge.next = mergeTwoLists(l1.next, l2);}else {Pmerge = l2;Pmerge.next = mergeTwoLists(l1, l2.next);}return Pmerge;}}

这时时间复杂度变为了：O(nKlogK)。优化不少性能。

0 0