def testDelete():
t = BST()
t.put(2)
t.put(1)
t.put(3)
t.delete(1)
assert not t.contains(1)
class Set():
def __init__(self):
self.bst = BST()
def add(self, e):
self.bst.add(e)
def contains(self, e):
return self.bst.contains(e)
def remove(self, e):
self.bst.remove(e)
#返回set中大于等于e的最小元素
def ceil(self, e):
pass
#返回set中小于等于e的最大元素
def floor(self, e):
pass
def getSize(self):
return self.bst.getSize()
def isEmpty(self):
return self.bst.isEmpty()
class BSTSet(SetBase):
def __init__(self):
self._bst = BST()
def get_size(self):
return self._bst.size()
def is_empty(self):
return self._bst.is_empty()
def add(self, e):
return self._bst.add(e)
def contains(self, e):
return self._bst.contains(e)
def remove(self, e):
return self._bst.remove(e)
class BSTSet(Set):
def __init__(self):
self.bst = BST()
# 判断集合是否为空
def is_empty(self):
return self.bst.is_empty()
# 添加 ele 至集合
def add(self, ele):
self.bst.add(ele)
# 是否包含元素 ele
def contains(self, ele):
return self.bst.contains(ele)
# 从集合内删除 ele
def remove(self, ele):
self.remove(ele)
# 获取集合大小
def get_size(self):
return self.bst.get_size()
f.close()
f = open('names_2.txt', 'r')
names_2 = f.read().split("\n") # List containing 10000 names
f.close()
duplicates = []
# this solution is 0(n ^ 2)
BST = BST('names')
for name1 in names_1:
BST.insert(name1)
for name2 in names_2:
if BST.contains(name2):
duplicates.append(name2)
end_time = time.time()
print(f"{len(duplicates)} duplicates:\n\n{', '.join(duplicates)}\n\n")
print(f"runtime: {end_time - start_time} seconds")
# ---------- Stretch Goal -----------
# Python has built-in tools that allow for a very efficient approach to this problem
# What's the best time you can accomplish with no restrictions on techniques or data
# structures?
# This solution is 0(n) !!!!!!!!!!!!!!!!!!!!!!!!!!!!!
n1 = set(names_1)
n2 = set(names_2)
#!/usr/env/bin python3
# File that uses BST
from BST import BST
bst = BST(25)
bst.insert(12)
bst.insert(50)
bst.insert(6)
bst.insert(45)
bst.insert(1)
print("Does tree contain 1 {0}".format(bst.contains(1)))
print("Does tree contain 5 {0}".format(bst.contains(5)))
print("Does tree contain 12 {0}".format(bst.contains(12)))
print("Does tree contain 51 {0}".format(bst.contains(51)))
def testPut():
t = BST()
t.put(2)
t.put(1)
t.put(3)
assert t.contains(1)
def testInit():
t = BST()
assert t.contains(1) == False
