Пример #1
0
def test_popping_from_stack_removes_item():
    stack = Stack()
    stack.push(1)

    assert_equal(stack.empty(), False)
    assert_equal(stack.pop(), 1)
    assert_equal(stack.empty(), True)
Пример #2
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def main():
    print "\ncheck stack"
    stack = Stack(1, 2, 34, 5)
    for x in range(0, 5):
        stack.push(x)
    print stack
    for x in range(0, 15):
        print "".join(["size=", str(len(stack)), " cur_node=", str(stack.pop())])

    print "\ncheck queue"
    queue = Queue(1, 2, 34, 5)
    for x in range(0, 5):
        queue.enter(x)
    print stack
    for x in range(0, 15):
        print "".join(["size=", str(len(queue)), " cur_node=", str(queue.exit())])

    print "\ncheck BSTree"
    tree = BSTree(1, 2, 34, 5)
    print tree
    print tree.find(10)
    print tree.find(5)
    print tree.max()
    print tree.min()
    print tree.successor(34)
    print tree.successor(5)
    print tree.predecessor(1)
    print tree.predecessor(2)
Пример #3
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def test_popping_from_stack_returns_last_item():
    stack = Stack()
    stack.push(1)
    stack.push(2)
    stack.push(3)

    assert_equal(stack.pop(), 3)
Пример #4
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    def midorder(self, f):
        """
        B树中序遍历
        :param f:
        :return:
        """
        result = []
        stack = Stack()
        cur_node = self.__root
        if cur_node.is_leaf:
            return map(f, cur_node.keys)

        while True:
            if cur_node.is_leaf:
                # 到叶节点了,开始把叶节点的所有关键字都遍历掉
                result.extend(map(f, cur_node.keys))
                # 开始从栈中取元素,遍历下一个节点叶节点
                if stack.empty():
                    return result
                cur_node, i = stack.pop()
                result.append(f(cur_node.keys[i]))
                if i < len(cur_node) - 1:
                    stack.push((cur_node, i + 1))
                cur_node = cur_node.childs[i + 1]
            else:
                stack.push((cur_node, 0))
                cur_node = cur_node.childs[0]
        return result
Пример #5
0
Файл: 2.py Проект: mhluska/ctci5
class MinStack:
  def __init__(self):
    self.mins  = Stack()
    self.items = Stack()

  def push(self, item):
    min = self.mins.peek()
    if min == None or item < min:
      self.mins.push(item)

    return self.items.push(item)

  def pop(self):
    item = self.items.pop()
    if item == self.mins.peek():
      self.mins.pop()

    return item

  def min(self):
    return self.mins.peek()
Пример #6
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def sort_stack(stack):
    sorted_stack = stack
    stack = Stack()

    while not sorted_stack.is_empty():
        stack.push(sorted_stack.pop())

    while not stack.is_empty():
        tmp = stack.pop()
        while not sorted_stack.is_empty() and sorted_stack.peek() < tmp:
            stack.push(sorted_stack.pop())
        sorted_stack.push(tmp)
Пример #7
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    def remove_even(self):
        tstack = Stack()

        # empty the stack and store even numbers in
        # another, temporary stack (in reverse order)
        while not self.empty:
            popped = self.pop()
            if popped % 2 == 1:
                tstack.push(popped)

        # move the data from temporary stack back to original
        while not tstack.empty:
            self.push(tstack.pop())
Пример #8
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 def preorder(self, f):
     result = []
     stack = Stack(self.__root)
     while True:
         cur_node = stack.pop()
         # 栈中没有元素的时候就表示所有的元素都已经遍历完了
         if cur_node is None:
             break
         result.append(f(cur_node.value))
         if cur_node.left is not None:
             stack.push(cur_node.left)
         if cur_node.right is not None:
             stack.push(cur_node.right)
     return result
Пример #9
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class TestStack(unittest.TestCase):
    def setUp(self):
        # Create stacks for the tests to use
        self.new = Stack()
        self.empty = Stack()
        self.empty.push('hi')
        self.empty.pop()
        # Don't add in ascending or descending order
        self.non_empty = Stack()
        self.non_empty.push(5)
        self.non_empty.push(2)
        self.non_empty.push(7)
        self.non_empty.push(2)

    def test_length(self):
        self.assertEqual(len(self.new), 0)
        self.assertEqual(len(self.empty), 0)
        self.assertEqual(len(self.non_empty), 4)

    def test_is_empty(self):
        self.assertTrue(self.new.is_empty())
        self.assertTrue(self.empty.is_empty())
        self.assertFalse(self.non_empty.is_empty())

    def test_lifo_order(self):
        self.assertEqual(self.non_empty.pop(), 2)
        self.assertEqual(self.non_empty.pop(), 7)
        self.assertEqual(self.non_empty.pop(), 2)
        self.assertEqual(self.non_empty.pop(), 5)

    def test_access_to_empty(self):
        with self.assertRaises(AssertionError):
            self.new.top()
        with self.assertRaises(AssertionError):
            self.empty.top()
        with self.assertRaises(AssertionError):
            self.new.pop()
        with self.assertRaises(AssertionError):
            self.empty.pop()

    def test_membership(self):
        self.assertFalse(2 in self.new)
        self.assertFalse(2 in self.empty)
        self.assertTrue(2 in self.non_empty)
        self.assertTrue(5 in self.non_empty)
        self.assertTrue(7 in self.non_empty)
Пример #10
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 def midorder(self, f):
     result = []
     stack = Stack(self.__root)
     cur_node = self.__root.left
     # 第一个阶段首先把所有树左边的节点放进栈里,这个时候并不遍历
     # 第二个阶段的时候由于左节点遍历了之后,再遍历右节点
     while not stack.empty() or cur_node is not None:
         # 第二个判断条件比较重要,因为如果根节点没有左子树,这个时候栈就是空的,会直接退出循环
         if cur_node is not None:
             stack.push(cur_node)
             cur_node = cur_node.left
         else:
             cur_node = stack.pop()
             result.append(f(cur_node.value))
             cur_node = cur_node.right
     return result
Пример #11
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def base_converter(dec_number, base):
    digits = "0123456789ABCDEF"
    s = Stack()
    n = dec_number
    while n > 0:
        number = n % base
        if number > 9:
            s.push(digits[number])
        else:
            s.push(n % base)
        n = n // base
    number = s.size()
    string = ''
    for i in range(number):
        string = string + str(s.pop())

    return string
Пример #12
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 def midorder(self, f):
     """
     中序遍历
     :param f:访问一个节点的时候要对节点进行处理的函数
     :return:
     """
     result = []
     stack = Stack(self.__root)
     cur_node = self.__root.left
     # 第一个阶段首先把所有树左边的节点放进栈里,这个时候并不遍历
     # 第二个阶段的时候由于左节点遍历了之后,再遍历右节点
     while not stack.empty() or cur_node is not self.Nil:
         # 第二个判断条件比较重要,因为如果根节点没有左子树,这个时候栈就是空的,会直接退出循环
         if cur_node is not self.Nil:
             stack.push(cur_node)
             cur_node = cur_node.left
         else:
             cur_node = stack.pop()
             result.append(f(cur_node))
             cur_node = cur_node.right
     return result
Пример #13
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    def dfs(self, gray_func, black_func):
        """
        图的深度遍历
        :param gray_func:
        :param black_func:
        :return:
        """
        gray_list = []
        black_list = []

        # 初始化
        for key in self.__dict.iterkeys():
            key.start_time = None
            key.end_time = None
            key.set_white()

        # 开始遍历
        counter = 0
        for key in self.__dict.iterkeys():
            if key.is_white():
                dfs_stack = Stack()
                key.set_gray()
                key.start_time = counter
                counter += 1
                dfs_stack.push(key)
                while not dfs_stack.empty():
                    cur_node = dfs_stack.pop()
                    gray_list.append(gray_func(key))
                    for end_node in self.__dict[cur_node]:
                        if end_node.is_white():
                            end_node.set_gray()
                            end_node.start_time = counter
                            counter += 1
                            dfs_stack.push(end_node)
                    cur_node.set_black()
                    black_list.append(black_func(cur_node))
                    cur_node.end_time = counter
                    counter += 1
        return gray_list, black_list
Пример #14
0
Файл: 3.py Проект: mhluska/ctci5
class SetOfStacks:
  def __init__(self, capacity):
    self.capacity = capacity
    self.stacks = [Stack()]
    self.active_stack_index = 0
    self.active_stack = self.stacks[0]

  def update_active_stack_next(self):
    if self.active_stack.size < self.capacity:
      return

    self.active_stack = Stack()
    self.stacks.append(self.active_stack)
    self.active_stack_index += 1

  def update_active_stack_prev(self):
    if self.active_stack.size > 0:
      return

    if self.active_stack_index == 0:
      return

    self.active_stack_index -= 1
    self.active_stack = self.stacks[self.active_stack_index]
    self.update_active_stack_prev()

  def push(self, data):
    self.update_active_stack_next()
    self.active_stack.push(data)
    return data

  def pop(self):
    self.update_active_stack_prev()
    return self.active_stack.pop()

  def pop_at(self, index):
    stack = self.stacks[index]
    if stack:
      return stack.pop()
Пример #15
0
def test_popping_from_empty_stack_raises_custom_error_message():
    stack = Stack()

    assert_equal(stack.pop(), "Error: Stack is empty")