leetcode: Basic Calculator

最新推荐文章于 2021-02-18 15:49:54 发布

martin_liang

最新推荐文章于 2021-02-18 15:49:54 发布

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分类专栏： C++/C 算法 leetcode

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本文将介绍如何实现一个基本的计算器，用于评估包含加减运算、括号和整数的简单表达式字符串。代码示例展示了如何通过遍历输入字符串并使用栈来处理各种操作符和括号。

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Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces.

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.

class Solution {
public:
    int calculate(string s) {
        
        stack<int> signs;
        
        int lastSign = 1;
        int curNum = 0;
        int retSum = 0;
        
        for (auto c: s)
        {
            if (c == '(')
            {
                signs.push(lastSign);//进栈主要用在有括号的情况，如果没有括号，就不需要有栈
            }
            else if (c == ')')
            {
                signs.pop();
            }
            else if (c >= '0' && c <= '9')
            {
                curNum = curNum * 10 + (c-'0');
            }
            else if (c == '+' || c == '-')
            {
                retSum = retSum + lastSign * curNum;
                lastSign = (c == '+'? 1:-1);
                if (signs.size() > 0)
                    lastSign = lastSign * signs.top();
                curNum = 0;
            }
        }
        
        return retSum + lastSign * curNum;
    }
};