LeetCode023. Merge k Sorted Lists

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

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

---

第一想法很简单，把lists中所有value存进一个列表中，再对列表进行排序，然后根据排序重新构造新的链表。这种方法速度很快，内存占用较多，代码如下：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        newlist=[]
        for list in lists:
            while list:
                newlist.append(list.val)
                list = list.next
        newlist.sort()
        
        head = p = ListNode(0)
        for node in newlist:
            p.next = ListNode(node)
            p = p.next
            
        return head.next
            

同时也学习了python的heapq库函数，有关于堆，可以进行自动排序，不过这样速度会更慢，内存占用基本没有变化。关于heapq函数使用，参考https://www.cnblogs.com/bonelee/p/10090727.html

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        import heapq
        
        heap=[]
        for linklist in lists:
            while linklist:
                heapq.heappush(heap, linklist.val)
                linklist = linklist.next
        
        head = p = ListNode(0)
        
        for i in range(len(heap)):
            p.next = ListNode(heapq.heappop(heap))
            p = p.next
        return head.next