题目：
给你一个链表数组，每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中，返回合并后的链表。
示例 1：
输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2：
输入：lists = []
输出：[]
示例 3：
输入：lists = [[]]
输出：[]
提示：
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4
链接：https://leetcode-cn.com/problems/merge-k-sorted-lists
思路：
1、合并K个升序链表第1步是构建一个函数合并2个升序链表
2、分治法将合并K个升序链表逐个进行合并
Python代码：
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def merge2List2(self, list1, list2):
if not list1:
return list2
if not list2:
return list1
guard = ListNode(0)
head = guard
while (list1 and list2):
if list1.val<=list2.val:
guard.next = ListNode(list1.val)
list1 = list1.next
else:
guard.next = ListNode(list2.val)
list2 = list2.next
guard = guard.next
guard.next = list1 if list1 else list2
return head.next
def merge(self, lists, left, right):
if left == right:
return lists[left]
mid = left + (right-left)/2
l1 = self.merge(lists, left, mid)
l2 = self.merge(lists, mid+1, right)
return self.merge2List2(l1, l2)
def mergeKLists(self, lists):
"""
:type lists: List[ListNode]
:rtype: ListNode
"""
if not lists:
return
left = 0
right = len(lists)-1
return self.merge(lists, left, right)
Java代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode merge2Lists(ListNode l1, ListNode l2){
if (l1==null){
return l2;
}
if (l2==null){
return l1;
}
ListNode guard = new ListNode(0);
ListNode head = guard;
while(l1!=null && l2!=null){
if(l1.val <= l2.val){
head.next = new ListNode(l1.val);
l1 = l1.next;
}else{
head.next = new ListNode(l2.val);
l2 = l2.next;
}
head = head.next;
}
if(l1!=null){
head.next = l1;
}else{
head.next = l2;
}
return guard.next;
}
public ListNode merge(ListNode[] lists, int left, int right){
if(left==right){
return lists[left];
}
int mid = left + (right-left)/2;
ListNode l1 = merge(lists, left, mid);
ListNode l2 = merge(lists, mid+1, right);
return merge2Lists(l1, l2);
}
public ListNode mergeKLists(ListNode[] lists) {
if (lists.length==0){
return null;
}
int left=0;
int right = lists.length-1;
return merge(lists, left, right);
}
}
