leetcode-hard-array- 227. Basic Calculator II
mycode 29.58%
class Solution(object):
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
def deal(data,flag):
data[:] = data[::-1]
while data:
if len(data) == 1:
break
a = data.pop()
if a == '+':
b = data.pop()
c = last + b
data.append(c)
elif a == '-':
b = data.pop()
c = last - b
data.append(c)
else:
last = a
return data[0] data = []
s = s.strip()
tokens = ['*','/','+','-']
l , r = 0, 0
for i in s:
if not i : continue
if i not in tokens:
r += 1
else:
data.append(int(s[l:r]))
r += 1
l = r
data.append(i)
data.append(int(s[l:r])) if '*' not in data and "/" not in data:
return deal(data,0) res = []
data[:] = data[::-1]
while data:
if '*' not in data and "/" not in data:
break
a = data.pop()
if a == '*':
b = data.pop()
res.pop()
c = last*b
data.append(c)
elif a == '/':
b = data.pop()
res.pop()
if b == 0:
return None
c = last // b
data.append(c)
else:
last = a
res.append(a)
return deal(res + data[::-1],1)
参考
import math
class Solution(object): def apply_pending_op(self, stack, pending_op, cur_int):
if pending_op is None:
stack.append(cur_int)
elif pending_op == '-':
stack.append(-cur_int)
elif pending_op == '+':
stack.append(cur_int)
elif pending_op == '*':
left = stack.pop()
right = cur_int
stack.append(left * right)
elif pending_op == '/':
left = stack.pop()
right = cur_int
# bypasses integer division rounding toward negative infinity
quo = int(float(left) / right)
stack.append(int(quo))
else:
raise ValueError(pending_op) def calculate(self, s):
"""
:type s: str
:rtype: int
"""
cur_int = 0
stack = []
pending_op = None
for c in s:
if c.isdigit():
cur_int = cur_int * 10 + int(c)
elif c in ('*', '/', '+', '-'):
self.apply_pending_op(stack, pending_op, cur_int)
cur_int, pending_op = 0, c self.apply_pending_op(stack, pending_op, cur_int) return sum(stack)
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