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algorithms-python/algorithms/stack/switch_pairs.py at master · Mu-L/algorithms-python · GitHub
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algorithms-python
/
algorithms
/
stack
/
switch_pairs.py
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algorithms-python
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algorithms
/
stack
/
switch_pairs.py
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"""
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
"""
import
collections
def
first_switch_pairs
(
stack
):
storage_stack
=
[]
for
i
in
range
(
len
(
stack
)):
storage_stack
.
append
(
stack
.
pop
())
for
i
in
range
(
len
(
storage_stack
)):
if
len
(
storage_stack
)
==
0
:
break
first
=
storage_stack
.
pop
()
if
len
(
storage_stack
)
==
0
:
# case: odd number of values in stack
stack
.
append
(
first
)
break
second
=
storage_stack
.
pop
()
stack
.
append
(
second
)
stack
.
append
(
first
)
return
stack
def
second_switch_pairs
(
stack
):
q
=
collections
.
deque
()
# Put all values into queue from stack
for
i
in
range
(
len
(
stack
)):
q
.
append
(
stack
.
pop
())
# Put values back into stack from queue
for
i
in
range
(
len
(
q
)):
stack
.
append
(
q
.
pop
())
# Now, stack is reverse, put all values into queue from stack
for
i
in
range
(
len
(
stack
)):
q
.
append
(
stack
.
pop
())
# Swap pairs by appending the 2nd value before appending 1st value
for
i
in
range
(
len
(
q
)):
if
len
(
q
)
==
0
:
break
first
=
q
.
pop
()
if
len
(
q
)
==
0
:
# case: odd number of values in stack
stack
.
append
(
first
)
break
second
=
q
.
pop
()
stack
.
append
(
second
)
stack
.
append
(
first
)
return
stack
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