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LC 0224 0227 0772 [H] [M] [H] Basic Calculator I II III
Code with Senpai edited this page Sep 13, 2022
·
44 revisions
I: +- and ()
II: +-*/
III: +-*/ and ()
so best to study III because it works for I + II + III
- num = num*10 + int(s[i]) to handle tenths place
- for addition/subtraction, just loop and add/subtract
- for multiplication and division, use a stack to get the previous value
- for parens () grouped expressions, use recursion while tracking the end pointer
int division should truncate (NOT floor) towards 0: use int() or math.trunc() (//
breaks with negative numbers)
>>> -3/2
-1.5
>>> int(-3/2)
-1
>>> -3//2
-2
>>> math.trunc(-3/2)
-1
vocab:
- operator: +, -, *, /
- operand: number value being operated on
'''
the correct answer of -3/2 is -1 because we want to truncate the negative answer towards 0 (round up) with int(x/y) or math.trunc(x/y)
if it were 3/2 it would be 1 because we want to truncate the postive answer towards 0 (round down)
>>> -3/2
-1.5
>>> -3//2
-2
>>> int(-3/2)
-1
>>> int(3//2)
1
>>> int(-3//2)
-2
>>> math.trunc(-3/2)
-1
'''
ops = {
'+': lambda x: stack.append( x ),
'-': lambda x: stack.append( -x ),
'*': lambda x: stack.append( stack.pop()*x ),
'/': lambda x: stack.append( int( stack.pop()/x ) ) # math.trunc() also works
}
class Solution:
def calculate(self, s):
stack = []
num = 0
prev_op = '+'
for i, x in enumerate(s):
# print(i, x) # debug
if x.isdigit():
num *= 10 # handle tens places as we build out the current number with x
num += int(x) # add the actual number
# print(i, x, num) # debug
elif x in ops: # else doesn't work because there can be ' '
ops[prev_op](num) # op is actually the previous op and num is already calculated, we just push it on the stack
# print(stack) # debug
# reset
num = 0
prev_op = x
# if i == len(s) - 1: # if we're at the end, we still need to add the last op we calculated
ops[prev_op](num)
return sum(stack)
introduce j
as the end pointer for a grouped expression
the recursion of the overall function can return either just the answer sum(stack)
or the group too sum(stack), j
class Solution:
def calculate(self, s: str) -> int:
stack = []
num, prev_op = 0, '+'
i = 0 # we need to use a while loop here to track and manage i
while i < len(s):
# handle the usual
if s[i].isdigit():
num = num*10 + int(s[i])
elif s[i] in ops:
ops[prev_op](num)
num, prev_op = 0, s[i]
# handle groups
elif s[i] == '(': # for BC I and BC III
num, j = self.calculate(s[i+1:])
i += j # the recursive call returns and we move i past the grouped expression: (i+j)
elif s[i] == ')': # for BC I and BC III
ops[prev_op](num)
j = i+1 # j represents the end of the grouped expression after it is parsed because we moved i then add the +1 for ): )
return sum(stack), j
i += 1
ops[prev_op](num) # still need to do last operation
return sum(stack)
wrap it all in a helper method for recursion (better to stick to iterative, it is easier than this)
class Solution:
def calculate(self, s):
s += '$' # extra char to remove n-1 check
def helper(stack, i):
num, op = 0, '+'
while i < len(s): # while loop to control i
c = s[i]
if c.isspace():
i += 1
elif c.isdigit():
num = 10*num + int(c)
i += 1
elif c == '(':
num, i = helper([], i+1) # move past (
else: # '+-*/' or ')' or i == n-1
ops[op](num)
num, op = 0, c
i += 1
if c == ')':
return sum(stack), i # return ans to parent who invoked recursion
return sum(stack)
return helper([], 0)
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