如何实现一个只支持加减乘除的简单计算器II功能？

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实现一个基本的计算器，用于评估一个简单的表达式字符串。表达式字符串只包含非负整数、+、-、*、/运算符和空格。整数除法应截断为零。例如：输入：3 + 1

pythondef basic_calculator(expression): stack=[] num=0 sign='+'

for char in expression: if char.isdigit(): num=num * 10 + int(char) if char in '+-*/' or char==expression[-1]: if sign=='+': stack.append(num) elif sign=='-': stack.append(-num) elif sign=='*': stack.append(stack.pop() * num) elif sign=='/': stack.append(int(stack.pop() / num)) sign=char num=0

return sum(stack)

测试expression=3 + 1result=basic_calculator(expression)print(result) # 输出：4

Implement a basic calculator to evaluate a simple expression string.

The expression string contains onlynon-negativeintegers,+,-,*,/operators and empty spaces. The integer division should truncate toward zero.

Example 1:

Input: "3+2*2" Output: 7

Example 2:

Input: " 3/2 " Output: 1

Example 3:

Input: " 3+5 / 2 " Output: 5

Note:

• You may assume that the given expression is always valid.
• Do notuse theevalbuilt-in library function.

Approach #1: Stack. [Java]

class Solution { public int calculate(String s) { if (s == null || s.length() == 0) return 0; Stack<Integer> stack = new Stack(); s += ‘+‘; char op = ‘+‘; for (int i = 0, n = 0; i < s.length(); ++i) { char c = s.charAt(i); if (c >= ‘0‘ && c <= ‘9‘) { n = n * 10 + c - ‘0‘; continue; } if (c == ‘ ‘) continue; if (op == ‘+‘) stack.push(n); else if (op == ‘-‘) stack.push(-n); else if (op == ‘*‘) stack.push(stack.pop() * n); else if (op == ‘/‘) stack.push(stack.pop() / n); op = c; n = 0; } int ret = 0; while (!stack.empty()) ret += stack.pop(); return ret; } }