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RotateList.py
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RotateList.py
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"""
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
旋转链表。 k 非负。
k 超过链表的最大长度也可。
思路:
过一遍链表长度,k % length 取个模,防止 k 超级大。
之后 slow fast 两个,fast 先走 k 个,然后 slow 与 fast 同时走,走到最后 slow.next 即为从后到前 k 个的起点。
剩下的就是把原来的尾置换到前。
下面这个可以优化下,不过就测试来说已经可以了。
beat 100%
24ms ~ 36ms
测试地址:
https://leetcode.com/problems/rotate-list/description/
"""
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def rotateRight(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
if not k or not head:
return head
def getLength(node):
length = 0
while node:
node = node.next
length += 1
return length
length = getLength(head)
k = k % length
slow = head
fast = head
while k > 0:
fast = fast.next
k -= 1
while fast.next:
slow = slow.next
fast = fast.next
rotate_head = slow.next
if not rotate_head:
return head
slow.next = None
_rotate_head = rotate_head
while _rotate_head.next:
_rotate_head = _rotate_head.next
_rotate_head.next = head
return rotate_head