LeetCode——Merge k Sorted Lists

Discription：

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

Subscribe to see which companies asked this question.

思路：其实就是归并排序的最后一步归并操作。思想是递归分治，先把一个大问题分成2个子问题，然后对2个子问题的解进行合并。经过一次遍历就能找出已经有序的序列。就算是题目中给出的条件，已经有K个已经排好序的链表。然后递归的把问题分成2个，4个。。。。logK个子序列。然后每两个子序列的解进行合并。最后得到一个排好序的链表。时间复杂度为O(nlogK)。

代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {    public ListNode merge2Lists(ListNode list1, ListNode list2) {        ListNode head = new ListNode(0);
        ListNode cur = head;        while(list1 != null && list2 != null) {
            if(list1.val < list2.val) {
                cur.next = list1;
                list1 = list1.next;
            }
            else {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }        while(list1 != null) {
            cur.next = list1;
            list1 = list1.next;
            cur = cur.next;
        }        while(list2 != null) {
            cur.next = list2;
            list2 = list2.next;
            cur = cur.next;
        }        return head.next;
    }    public ListNode mergeKLists(ListNode[] lists) {        if(lists == null || lists.length == 0) {
            return null;
        }        if(lists.length == 1) {
            return lists[0];
        }        int mid = lists.length / 2;        ListNode list1 = mergeKLists(Arrays.copyOfRange(lists, 0, mid));
        ListNode list2 = mergeKLists(Arrays.copyOfRange(lists, mid, lists.length));        return merge2Lists(list1, list2);
    }
}

为何这道题Java的效率这么高，把C/C++都完爆了。