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linked list algorithms in python
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jerryderry committed Oct 5, 2018
1 parent b86cd91 commit a04b592
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92 changes: 92 additions & 0 deletions python/07_linkedlist/linked_list_algo.py
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"""
1) Reverse singly-linked list
2) Detect cycle in a list
3) Merge two sorted lists
4) Remove nth node from the end
5) Find middle node
Author: Wenru
"""

from typing import Optional


class Node:

def __init__(self, data: int, next=None):
self.data = data
self._next = next

# Reverse singly-linked list
# 单链表反转
# Note that the input is assumed to be a Node, not a linked list.
def reverse(head: Node) -> Optional[Node]:
reversed_head = None
current = head
while current:
reversed_head, reversed_head._next, current = current, reversed_head, current._next
return reversed_head

# Detect cycle in a list
# 检测环
def has_cycle(head: Node) -> bool:
slow, fast = head, head
while fast and fast._next:
slow = slow._next
fast = fast._next._next
if slow == fast:
return True
return False

# Merge two sorted linked list
# 有序链表合并
def merge_sorted_list(l1: Node, l2: Node) -> Optional[Node]:
if l1 and l2:
p1, p2 = l1, l2
fake_head = Node(None)
current = fake_head
while p1 and p2:
if p1.data <= p2.data:
current._next = p1
p1 = p1._next
else:
current._next = p2
p2 = p2._next
current = current._next
current._next = p1 if p1 else p2
return fake_head._next
return p1 or p2

# Remove nth node from the end
# 删除倒数第n个节点。假设n大于0
def remove_nth_from_end(head: Node, n: int) -> Optional[Node]:
fast = head
count = 0
while fast and count < n:
fast = fast._next
count += 1
if not fast and count < n: # not that many nodes
return head
if not fast and count == n:
return head._next

slow = head
while fast._next:
fast, slow = fast._next, slow._next
slow._next = slow._next._next
return head

def find_middle_node(head: Node) -> Optional[Node]:
slow, fast = head, head
fast = fast._next if fast else None
while fast and fast._next:
slow, fast = slow._next, fast._next._next
return slow

def print_all(head: Node):
nums = []
current = head
while current:
nums.append(current.data)
current = current._next
print("->".join(str(num) for num in nums))

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