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61. Rotate List
题目描述
Given a linked list, rotate the list to the right by k places, where k is non-negative.
例子
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
解法
- 第一次遍历,记录链表长度size
- 有k%size个结点翻转到了前面
解法
记录长度,两次遍历 - (先把链表连成环)
# 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 head or not head.next:
return head
# Mark size and tail
size = 1
tail = head
while tail.next:
tail = tail.next
size += 1
tail.next = head
# Rotate
k = k % size
p = head
for _ in range(size-k-1):
p = p.next
node = p.next
p.next = None
return node
143. Reorder List
题目描述
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You may not modify the values in the list’s nodes, only nodes itself may be changed.
例子
Example 1:
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Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
解法
法1 - split得到后半段(通过快慢指针)→反转后半段→merge
法2 - 建立辅助数组,存储所有结点后再操作
解法1
时间复杂度O(n),空间复杂度O(1)
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reorderList(self, head):
"""
:type head: ListNode
:rtype: void Do not return anything, modify head in-place instead.
"""
if not head or not head.next:
return
# Split - find the 2nd-half
slow = fast = head
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
node = slow.next
slow.next = None
# Reverse the 2nd-half
node = self.reverseList(node)
# Merge head and node
p = head
while p.next:
temp = node.next
node.next = p.next
p.next = node
p = node.next
node = temp
p.next = node
def reverseList(self, head):
if not head or not head.next:
return head
node = self.reverseList(head.next)
head.next.next = head
head.next = None
return node
解法2
时间复杂度O(n),空间复杂度O(n)
class Solution(object):
def reorderList(self, head):
"""
:type head: ListNode
:rtype: void Do not return anything, modify head in-place instead.
"""
arr = []
while head:
arr.append(head)
head = head.next
size = len(arr)
for i in range(size//2):
arr[i].next = arr[-i]
arr[-i].next = arr[i+1]
arr[size//2+1].next = None