Composition
Linked List
用Python实现链表 -
每个节点由first与rest组成,前者表示值,后者表示链接的剩余链表
- 将每个节点当作一堆二元组看待 -
最后一个节点指向空链表Link.empty,使用类属性表示 -
构造方式:Link(3, Link(4, Link(5, Link.empty))) ###
创建链表
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class Link:
# 使用空元组实现空节点,类属性
empty = ()
def __init__(self, first, rest=empty):
# 判断rest是否是空节点或是一个Link类
assert rest is Link.empty or isinstance(rest, Link)
self.first = first
self.rest = rest
s = Link(3, Link(4, Link(5)))
print(s.first)
print(s.rest.first)
print(s.rest.rest.first)
print(s.rest.rest.rest is Link.empty)操作列表
实现如下行为:range, map,
filter 我们使用递归实现
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class Link:
# 使用空元组实现空节点,类属性
empty = ()
def __init__(self, first, rest=empty):
# 判断rest是否是空节点或是一个Link类
assert rest is Link.empty or isinstance(rest, Link)
self.first = first
self.rest = rest
def range_link(start, end):
if start>=end:
return Link.empty
else:
return Link(start, range_link(start+1, end))
def map_link(f, s):
if s is Link.empty:
return s
else:
return Link(f(s.first), map_link(f, s.rest))
def filter_link(f, s):
if s is Link.empty:
return s
filter_rest = filter(f, s.rest)
if f(s.first):
return Link(s.first, filter_rest)
else:
return filter_rest链表的改变
可以通过对属性赋值来改变链表的first和rest属性
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>>> s = Link(1, Link(2, Link(3)))
>>> s.first = 5
>>> t = s.rest
>>> t.rest = s;
>>> s.first
5
>>> s.rest.rest.rest.rest.rest.first
2此赋值将s的后段与前段相接,创建了一个循环链表
插入节点
向一个有序链表中添加节点,保证链表仍有序
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def add(s, v):
assert s is not Link.empty
if s.first > v:
s.first, s.rest = v, Link(s.first, s.rest)
elif s.first < v and s.rest is Link.empty:
s.rest = Link(v)
elif s.first < v:
add(s.rest, v)
return sTree Class
使用类实现树 ### 创建树
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class Tree:
def __init__(self, label, branches = []):
self.label = label
for branch in branches:
assert isinstance(branch, Tree)
self.branches = list(branches)
def __repr__(self):
if self.branches:
branches_str = ', '+ repr(self.branches)
else:
branch_str = ''
return 'Tree({0}{1})'.format(repr(self.label), branch_str)
def __str__(self):
return '\n'.join(self.indented())
def indented(self):
lines = []
for b in self.branches:
for line in b.indented():
lines.append(' '+line)
return [str(self.label)] + lines
def is_leaf(self):
return not self.branches实现一些操作
如下代码实现了创建斐波那契树以及输出叶子节点
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def fib_tree(n):
if n==0 or n==1:
return Tree(n)
else:
left = fib_tree(n-2)
right = fib_tree(n-1)
fib_n = left.label + right.label
return Tree(fib_n, [left, right])
def leaves(t):
if t.is_leaf():
return [t.label]
else:
all_leaves = []
for b in t.branches:
all_leaves.extend(leaves(b))
return all_leaves剪枝
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# 通过每次重新构建branches列表,将等于n的值排除在列表外以实现
def prune(t, n):
t.branches = [b for b in t.branches if b.label!=n]
for b in t.branches:
prune(b, n)