LeetCode 224 Basic Calculator

LeetCode

Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:

Input: s = “1 + 1”
Output: 2

Example 2:

Input: s = “ 2-1 + 2 “
Output: 3

Example 3:

Input: s = “(1+(4+5+2)-3)+(6+8)”
Output: 23

Constraints:

`1 <= s.length <= 3 10<sup>5</sup>*sconsists of digits,‘+’,‘-‘,‘(‘,‘)’, and‘ ‘. *srepresents a valid expression. *‘+’is **not** used as a unary operation (i.e.,“+1”and“+(2 + 3)”is invalid). *‘-‘could be used as a unary operation (i.e.,“-1”and“-(2 + 3)”` is valid).
There will be no two consecutive operators in the input. Every number and running calculation will fit in a signed 32-bit integer.

题目大意：

实现字符串加减，但有括号。

算法思路：

括号题优先考虑用Stack。这里Stack不能只存数，因为括号前可以是正负，所以将这个信息以+1或-1也压入栈（栈不能混合字符和数字）
所以用一个stack，num是一个数，res是括号内的累积结果。num在处理完每一个数都要重设，res和sign在处理完每个括号都要重设。

运用模板，代码中含五种情况：空格，运算符，数字，左右括号。左括号将res和sign入栈，右括号将res和sign出栈，计算结果存在res

LeetCode 224 Basic Calculator 括号加减法, 同一层括号内求和遇括号入栈
LeetCode 227 Basic Calculator II 加减乘除, 和的每一项入栈，方便出栈计乘除
LeetCode 772 Basic Calculator III 加减乘除括号, L227的递归版

注意事项：

1. 括号前可以是正负，所以将这个信息以+1或-1也压入栈
2. 左括号无论前面是正负都要入栈
3. num在处理完每一个数都要重设，res和sign在处理完每个括号都要重设

Python代码：

# 1-(2+3)+(4+5)
# 1+2+3
def calculate(self, s: str) -> int:
	res, num, stack, sign = 0, 0, [], 1
	s += '+'
	for char in s:
		if char == ' ':
			continue
		if char.isdigit():
			num = num * 10 + int(char)
		if char == '+':
			res += sign * num
			sign = 1
			num = 0
		if char == '-':
			res += sign * num
			sign = -1
			num = 0
		if char == '(':
			stack.append(res) # [-4+]
			stack.append(sign) #
			sign = 1
			res = 0
		if char == ')':
			res += sign * num # 9
			prev_sign = stack.pop() # +
			tmp = stack.pop() # -4
			res = tmp + prev_sign * res # -4 +9
			# sign = 1 next one will be +-, so num = 0 and sign doesn't matter
			num = 0
		# else:
		 #   res += char
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(n)