You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
Example 1:
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
1->4->5,
1->3->4,
2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6
Example 2:
Input: lists = []
Output: []
Example 3:
Input: lists = [[]]
Output: []
Constraints:
- k == lists.length
- 0 <= k <= 10^4
- 0 <= lists[i].length <= 500
- -10^4 <= lists[i][j] <= 10^4
- lists[i] is sorted in ascending order.
- The sum of lists[i].length won't exceed 10^4.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
return partition(lists,0,lists.length-1);
}
public ListNode partition(ListNode[] lists, int s ,int e){
if(s==e){
return lists[s];
}
if(s<e){
int q = (s+e)/2;
ListNode l1 = partition(lists, s ,q);
ListNode l2 = partition(lists, q+1, e);
return merge(l1,l2);
}else{
return null;
}
}
public static 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;
}else{
l2.next = merge(l1,l2.next);
return l2;
}
}
}使用二分法，否则会超时