Leetcode Basic Calculator I && II_i < len && character.isdigit(expression.charat(i))-优快云博客

本文链接：https://blog.youkuaiyun.com/sundawei2016/article/details/75453325

本文介绍了一种简单表达式计算器的实现方法，该计算器能够处理包含括号、加减乘除运算符及空格的非负整数表达式，并正确进行运算。文中提供了两个示例代码：一个用于处理加减法及括号的表达式；另一个则能处理包括乘除在内的四则运算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

    public int calculate(String s) {
        if (s==null || s.length() == 0) return 0;  
        Stack<Integer> st = new Stack<Integer>();  
        int res = 0, sign = 1, len = s.length();
        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                int cur = c - '0';
                while (i + 1 < len && Character.isDigit(s.charAt(i + 1))) {
                    cur = 10 * cur + s.charAt(++i) - '0';
                }
                res += sign * cur;
            }
            else if (c == '-') sign = -1;
            else if (c == '+') sign = 1;
            else if (c == '(') {
                st.push(res);
                res = 0;
                st.push(sign);
                sign = 1;
            }
            else if (c == ')') {
                res = st.pop() * res + st.pop();
                sign = 1;
            }
            
        }
        return res;

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

public int calculate(String s) {
        Stack<Integer> stack = new Stack<>();
        char sign = '+';
        int len = s.length();
        int n = 0;
        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                n = n * 10 + c - '0';
            }
            if (!Character.isDigit(c) && c != ' ' || i == len - 1) {
                if (sign == '+') {
                    stack.push(n);
                } else if (sign == '-') {
                    stack.push(-n);
                } else if (sign == '*') {
                    stack.push(stack.pop() * n);
                } else if (sign == '/') {
                    stack.push(stack.pop() / n);
                }
                sign = c;
                n = 0;
            }
        }
        int re = 0;
        while (!stack.isEmpty()) re += stack.pop();
        return re;
    }