class Solution:
def calculate(self, s: str) -> int:
# 初始化sign为 “+”，是因为开头是数字
num ,stack ,sign = 0 , [] , '+'
for i in range(len(s)):
ch = s[i]
if ch.isdigit():
num = num * 10 + int(ch)
#根据当前数字之前的符号，来决定如何处理当前数值
# 并将处理后的数值压入栈中
if ch in "+-*/" or i == len(s)-1:
if sign == "+" :
stack.append(num)
elif sign == "-" :
stack.append(-num)
elif sign == "*":
stack.append(stack.pop() * num)
else:
stack.append(int(stack.pop()/num))
num = 0
sign = ch
return sum(stack)- Topic analysis
- From elementary school knowledge, we can understand that in basic calculations, first multiply and divide, then add and subtract. If there are parentheses, first calculate the value in the parentheses.
- For this topic, there are no parentheses, so we first calculate multiplication and division, and then add and subtract
- Basically, this problem needs to use the data structure of the stack to solve this problem. Initialize a stack and initialize a symbol. The initial symbols are all plus signs.
- If it is a number, first calculate the string to an integer
- If the current string is a character, then if it is addition and subtraction, multiply num by its coefficient, and then push it to the stack; if it is multiplication and division, first pop out the stack and perform operations on the current element, and then push it to the stack
- Then update the current num to 0, and record the current coefficient
- Summarize
- One of the test points is how to convert a string into a decimal integer
- Test center two, what should be done when encountering'+-*/'
- Test point three, how to update the current coefficient and the next integer after each calculation
class Solution:
def calculate(self, s: str) -> int:
sDeque = deque(s)
def helper(sDeque):
num = 0
sign = "+"
stack = []
while len(sDeque) > 0:
# 注意 这里需要从头部开始pop
char = sDeque.popleft()
if char.isdigit():
num = num * 10 + int(char)
if char == '(':
#遇到左括号，开始递归调用
num = helper(sDeque)
if (not char.isdigit() and char != " ") or len(sDeque) == 0:
if sign == "+":
stack.append(num)
elif sign == "-":
stack.append(-num)
elif sign =='*':
stack.append(stack.pop()*num)
elif sign == '/':
stack.append(stack.pop()/num)
# 重置num和sign
num = 0
sign = char
if char == ")":
#遇到右括号直接结束当前的递归
break
return sum(stack)
return helper(sDeque)- Topic analysis
- It is an upgraded version of the previous topic, with brackets added.
- First convert the previous string into a deque
- Others remain unchanged. If we encounter a left parenthesis, then we need to perform calculations recursively. If we encounter a right parenthesis to end this recursive call, just jump out of the while loop.
Here is a post on how to dismantle a complicated problem: to implement a calculator , labuladong is definitely a big man, and the explanation of this topic is very clear and clear.
Summary: I thought the stack was quite simple, but found that the simpler the data structure is really not so simple in the application of specific problems.