def test_switch_pairs(self):
# Test case: even number of values in stack
# bottom [3, 8, 17, 9, 1, 10] top
self.assertEqual([8, 3, 9, 17, 10, 1],
first_switch_pairs([3, 8, 17, 9, 1, 10]))
self.assertEqual([8, 3, 9, 17, 10, 1],
second_switch_pairs([3, 8, 17, 9, 1, 10]))
# Test case: odd number of values in stack
# bottom [3, 8, 17, 9, 1] top
self.assertEqual([8, 3, 9, 17, 1], first_switch_pairs([3, 8, 17, 9,
1]))
self.assertEqual([8, 3, 9, 17, 1],
second_switch_pairs([3, 8, 17, 9, 1]))
def test_switch_pairs(self):
# Test case: even number of values in stack
# bottom [3, 8, 17, 9, 1, 10] top
self.assertEqual([8, 3, 9, 17, 10, 1],
first_switch_pairs([3, 8, 17, 9, 1, 10]))
self.assertEqual([8, 3, 9, 17, 10, 1],
second_switch_pairs([3, 8, 17, 9, 1, 10]))
# Test case: odd number of values in stack
# bottom [3, 8, 17, 9, 1] top
self.assertEqual([8, 3, 9, 17, 1],
first_switch_pairs([3, 8, 17, 9, 1]))
self.assertEqual([8, 3, 9, 17, 1],
second_switch_pairs([3, 8, 17, 9, 1]))
"""
Given a stack, switch_pairs function takes a stack as a parameter and that
switches successive pairs of numbers starting at the bottom of the stack.
For example, if the stack initially stores these values:
bottom [3, 8, 17, 9, 1, 10] top
Your function should switch the first pair (3, 8), the second pair (17, 9), ...:
bottom [8, 3, 9, 17, 10, 1] top
if there are an odd number of values in the stack, the value at the top of the
stack is not moved: For example:
bottom [3, 8, 17, 9, 1] top
It would again switch pairs of values, but the value at the top of the stack (1)
would not be moved
bottom [8, 3, 9, 17, 1] top
Note: There are 2 solutions:
first_switch_pairs: it uses a single stack as auxiliary storage
second_switch_pairs: it uses a single queue as auxiliary storage
"""
from algorithms.stack import first_switch_pairs, second_switch_pairs
a = [3, 8, 17, 9, 1, 10]
print(first_switch_pairs(a))
print(second_switch_pairs(a))