Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6
合并链表很简单，而且还是有序的，k个指针前进就行。写代码的时候只要随时记得保持良好习惯，尽量用少量的判断来包括多种条件进去，这样写出来的代码就不会和严蔚敏的数据结构书上一样丑了。
# 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 mergeKLists(self, lists):
"""
:type lists: List[ListNode]
:rtype: ListNode
"""
k_cursors = [i for i in lists if i ]
result = ListNode()
cursor = result
while len(k_cursors) > 0:
smallest = self.find_smallest_in_k_cursors(k_cursors)
new_node = ListNode(smallest)
cursor.next = new_node
cursor = cursor.next
result = result.next
return result
def find_smallest_in_k_cursors(self, k_cursors):
smallest = float("inf")
remeber_cursor = {}
for i, k_cursor in enumerate(k_cursors):
if k_cursors[i].val < smallest:
smallest = k_cursors[i].val
remeber_cursor[smallest] = i
if not k_cursors[remeber_cursor[smallest]].next:
k_cursors.pop(remeber_cursor[smallest])
else:
k_cursors[remeber_cursor[smallest]] = k_cursors[remeber_cursor[smallest]].next
return smallest
原文：https://www.cnblogs.com/importsober/p/13185923.html