https://leetcode.com/problems/merge-k-sorted-lists/
思路
一个简单的思路是把k个数组两两相加,但这样会超时,所以采用分治算法。合并两个数组的复杂度是O(n)的话,总的复杂度应该是O(nlog k)
AC代码
# 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
"""
return self.merge2Lists(lists)
def merge2Lists(self, lists):
if len(lists) == 0:
return None
if len(lists) == 1:
return lists[0]
if len(lists) > 2:
L = len(lists)
return self.merge2Lists([self.merge2Lists(lists[:L/2]), self.merge2Lists(lists[L/2:])])
l1, l2= lists
if l1 == None:
return l2
if l2 == None:
return l1
res = ListNode(0)
head = res
while(l1 and l2):
if l1.val < l2.val:
res.val = l1.val
res.next = ListNode(0)
res = res.next
l1 = l1.next
else:
res.val = l2.val
res.next = ListNode(0)
res = res.next
l2 = l2.next
if l1:
res.val = l1.val
res.next = l1.next
else:
res.val = l2.val
res.next = l2.next
return head