コード例 #1
0
 def test_pop(self):
     s = Stack(['A', 'B', 'C'])
     assert s.pop() == 'C'
     assert s.length() == 2
     assert s.pop() == 'B'
     assert s.length() == 1
     assert s.pop() == 'A'
     assert s.length() == 0
     assert s.is_empty() is True
     with self.assertRaises(ValueError):
         s.pop()
コード例 #2
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 def test_length(self):
     s = Stack()
     assert s.length() == 0
     s.push('A')
     assert s.length() == 1
     s.push('B')
     assert s.length() == 2
     s.pop()
     assert s.length() == 1
     s.pop()
     assert s.length() == 0
コード例 #3
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 def test_push(self):
     s = Stack()
     s.push('A')
     assert s.peek() == 'A'
     assert s.length() == 1
     s.push('B')
     assert s.peek() == 'B'
     assert s.length() == 2
     s.push('C')
     assert s.peek() == 'C'
     assert s.length() == 3
     assert s.is_empty() is False
コード例 #4
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ファイル: reverseInteger.py プロジェクト: ysawiris/SPD1.4
def reverse(nums):
    #declare your stack
    Stack = LinkedStack()  #init the stack

    #declare a variable to get track of the reversed number
    reversed = []

    #add each number in nums to the stack
    for num in nums:  # Loop through the whole list O(n)
        if num == '-':
            reversed.append(num)  #adds sign to list O(1)
        if num.isdigit():
            Stack.push(num)  #adds num to stack O(1)

    #traverse through the stack
    while Stack.length() > 0:  # Loops through the whole stack
        #add the head of the stack then delete the head
        reversed.append(Stack.peek())  # adds number back to list O(1)
        Stack.pop()  # deletes current stack to move to the next one O(1)

    if reversed[0] == '-':
        if reversed[1] == '0':
            reversed.pop(1)  #removes a digit O(1)

    if reversed[0] == '0':
        reversed.pop(0)  #removes a digit O(1)

    return "".join(reversed)
コード例 #5
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    def _traverse_in_order_iterative(self, node, visit):
        """Traverse this binary tree with iterative in-order traversal (DFS).
        Start at the given node and visit each node with the given function.
        Running time: O(n) since we visit every node
        Memory usage: O(logn) for the depth of the tree"""
        # -create an empty stack
        # -curr variable which starts off with the self.root (will be updated in a while loop!!)
        # -go in a while loop while true (will be false when the stack is empty)
        #   -call push() on curr while there's a value
        #   -update the curr to become the curr's left
        #   -once we hit a None value:
        #       - we'd want to pop() off the stack
        #       - append it to the list
        #       - update curr to be the recently popped node's right node
        #       - however, if the stack is empty in here, we set the while loop to false and exit the function!

        stack = LinkedStack()  # will act as the recursive call
        curr = self.root  # start at the root
        stillLoop = True  # loop until the stack is empty

        while stillLoop:
            # we're at the end of a leaf node
            if curr == None:
                if stack.length() > 0:
                    node = stack.pop()
                    visit(node.data)
                    curr = node.right
                # our stack is empty so we can finally return once all the nodes have been visited
                else:
                    stillLoop = False
            # we can still go down the left side
            else:
                stack.push(curr)
                curr = curr.left
コード例 #6
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 def test_init(self):
     s = Stack()
     assert s.peek() is None
     assert s.length() == 0
     assert s.is_empty() is True
コード例 #7
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 def test_init_with_list(self):
     s = Stack(['A', 'B', 'C'])
     assert s.peek() == 'C'
     assert s.length() == 3
     assert s.is_empty() is False