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
代码第二部分不理解,后续补充
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
//实际上是对链表的k个节点组成的head分别处理
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if(lists==null||lists.length==0){
return null;
}
return MSort(lists,0,lists.length -1);//lists.lengt=k,k个链表头
}
//这一段代码不怎么理解
public ListNode MSort(ListNode[] lists, int low, int high) {
if (low < high) {
int mid = (low + high) / 2;
ListNode leftlist = MSort(lists, low, mid);
ListNode rightlist = MSort(lists, mid + 1, high);
return mergeTwoLists(leftlist, rightlist);
}
// 如果相等,只有一个元素,返回即可
return lists[low];
}
public ListNode mergeTwoLists(ListNode l1,ListNode l2){
ListNode res= null;
if(l1==null)
return l2;
if(l2==null)
return l1;
if(l1.val<l2.val){
res=l1;
l1.next=mergeTwoLists(l1.next,l2);
}
else{
res=l2;
l2.next=mergeTwoLists(l1,l2.next);
}
return res;
}
}