def __init__(self):
super(double_queue_with_stack, self).__init__()
self.left = stack.stack()
self.right = stack.stack()
self.mid = stack.stack()
self.mid_type = True
self.len = 0
def prase_nfa(reg):
'''
@brief 使用nfa解析正则表达式
@param ref 模式串
@return
'''
stk = stack()
g = diagraph()
for i in range(len(reg)):
lp = i
if reg[i] == '(' or reg[i] == '|':
stk.push(i)
elif reg[i] == ')':
or_id = stk.top()
stk.pop()
if (reg[i] == '|'):
lp = stk.top()
g.add_edge(lp, or_id + 1)
g.add_edge(or_id, i)
else:
lp = or_id
if i < len(reg) - 1 and reg[i + 1] == '*':
g.add_edge(lp, i + 1)
g.add_edge(i + 1, lp)
if reg[i] == '(' or reg[i] == '*' or reg[i] == ')':
g.add_edge(i, i + 1)
return g
def in2post(line):
'''
@brief 1.3.10 中序表达式转换为后序表达式，不含括号的表达式转换
@param line 中序表达式
'''
op_stk = stack.stack()
ret_line = ''
for ch in line:
if ch.isdigit():
ret_line += ch
elif ch == '(':
op_stk.push(ch)
elif ch == ')':
tmp = op_stk.pop()
ret_line += tmp
while tmp != ')':
tmp = op_stk.pop()
else:
while get_prior(op_stk.top()) > get_prior(ch):
ret_line += op_stk.pop()
op_stk.push(ch)
while not op_stk.empty():
ret_line += op_stk.pop()
return op_stk
else:
def post_traversal_no_II(root, visit_func):
'''
@brief 后序遍历非递归
@param root 当前访问的结点
@param visit_func 访问结点的函数
'''
if root is None:
return
stk = stack.stack()
pre = None
cur = None
stk.push(root)
while not stk.empty():
cur = stk.top()
if pre is None or pre.left == cur or pre.right == cur:
if cur.left is not None:
stk.push(cur.left)
elif cur.right is not None:
stk.push(cur.right)
elif cur.left == pre:
if cur.right is not None:
stk.push(cur.right)
else:
visit_func(cur)
stk.pop()
pre = cur
def complete_left_brackets(line):
'''
@brief 1.3.9 补全表达式中的左括号,表达式中只有右括号而没有左括号，将左括号补全
@param line 表达式
@return 补全后的表达式字符串
'''
val_stk = stack.stack()
op_stk = stack.stack()
for ch in line:
if ch.isdigit():
val_stk.push(ch)
elif ch in '+-*/':
op_stk.push(ch)
elif ch == ')':
cur_value = '(' + op_stk.pop() + val_stk.pop() + ch
val_stk.push(cur_value)
else:
raise Exception('bad character')
return val_stk.pop()
def bracket_match(line):
'''
@brief 括号匹配，支持(),[],{}，对于不是相关字符接略过
@param line 括号字符串
'''
stk = stack.stack()
for ch in line:
if is_left_bracket(ch):
stk.push(ch)
elif is_right_bracket(ch):
if not match_bracket(ch, stk.pop()):
return False
else:
continue
def post_traversal_no(root, visit_func):
'''
@brief 后序遍历非递归
@param root 当前访问的结点
@param visit_func 访问结点的函数
'''
rst = stack.stack()
snd = stack.stack()
while root is not None or not rst.empty():
while root is not None:
rst.push(root)
snd.push(root)
root = root.right
if not rst.empty():
root = rst.top()
rst.pop()
root = root.left
while not snd.empty():
root = snd.top();
snd.pop()
visit_func(root)
def quick_sort_cyc(l, start, end, hook_func=None):
'''
@brief 快速排序非递归版本
@param l list
@param start
@param end
@parma hook_func
@note 建立一个容器，容器的每个元素是一个tuple表示需要进行partition的start和end
'''
rst = stack.stack()
snd = stack.stack()
rst.push((start, end))
while not rst.empty():
while not rst.empty():
start, end = rst.pop()
if end < start:
continue
anchor = partition(l, start, end, hook_func)
snd.push((start, anchor - 1))
snd.push((anchor + 1, end))
rst, snd = snd, rst
return l
def in_traversal_no(root, visit_func):
'''
@brief 中序遍历非递归版本
@param root 当前访问的结点
@param visit_func 访问结点的函数
'''
stk = stack.stack()
index = root
while not stk.empty() or index is not None:
while index is not None:
stk.push(index)
index = index.left
index = stk.top()
visit_func(index)
stk.pop()
index = index.right
def select(self, node, k):
'''
@brief 返回排名为k的节点
@note 使用中序遍历的思路
'''
stk = stack.stack()
while not stk.empty() and node is None and k != 0:
while node is not None:
stk.push(node)
node = node.left
if not stk.empty():
node = stk.pop()
k -= 1
node = node.right
return node
def pre_traversal_no(root, visit_func):
'''
@brief 先序遍历非递归
@param root 当前访问的结点
@param visit_func 访问结点的函数
'''
stk = stack.stack()
index = root
while index is not None and not stk.empty():
while index is not None:
visit_func(index)
stk.push(index)
index = index.left
if not stk.empty():
index = stk.top()
stk.pop()
index = index.right
def rank(self, node, key):
'''
@brief 返回node中的子树种小于key键的数量
'''
stk = stack.stack()
no = 0
while not stk.empty() and node is None:
while node is not None:
stk.push(node)
node = node.left
if not stk.empty():
node = stk.pop()
if node.key < key:
no += 1
else:
return no
node = node.right
return no
def __init__(self):
self.rst = stack.stack()
self.snd = stack.stack()
