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split-linked-list-in-parts.py
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split-linked-list-in-parts.py
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# Time: O(n + k)
# Space: O(1)
# Given a (singly) linked list with head node root,
# write a function to split the linked list into k consecutive linked list "parts".
#
# The length of each part should be as equal as possible:
# no two parts should have a size differing by more than 1.
# This may lead to some parts being null.
#
# The parts should be in order of occurrence in the input list,
# and parts occurring earlier should always have a size greater than or equal parts occurring later.
#
# Return a List of ListNode's representing the linked list parts that are formed.
#
# Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]
# Example 1:
# Input:
# root = [1, 2, 3], k = 5
# Output: [[1],[2],[3],[],[]]
# Explanation:
# The input and each element of the output are ListNodes, not arrays.
# For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3,
# and root.next.next.next = null.
# The first element output[0] has output[0].val = 1, output[0].next = null.
# The last element output[4] is null, but it's string representation as a ListNode is [].
#
# Example 2:
# Input:
# root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
# Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
# Explanation:
# The input has been split into consecutive parts with size difference at most 1,
# and earlier parts are a larger size than the later parts.
#
# Note:
# - The length of root will be in the range [0, 1000].
# - Each value of a node in the input will be an integer in the range [0, 999].
# - k will be an integer in the range [1, 50].
#
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def splitListToParts(self, root, k):
"""
:type root: ListNode
:type k: int
:rtype: List[ListNode]
"""
n = 0
curr = root
while curr:
curr = curr.next
n += 1
width, remainder = divmod(n, k)
result = []
curr = root
for i in xrange(k):
head = curr
for j in xrange(width-1+int(i < remainder)):
if curr:
curr = curr.next
if curr:
curr.next, curr = None, curr.next
result.append(head)
return result