leetcode--23--合并K个升序链表

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);
    }
}