. Given a string of expression, a+b/c. Please use a stack to re-order the expression into abc/+.

可以使用栈来实现字符串的重新排列，具体步骤如下：

1. 遍历原始字符串，将每个字符依次入栈；
2. 当遇到斜杠“/”时，将栈中的元素全部出栈并组成新的字符串，再将斜杠“/”添加到新字符串的末尾；
3. 最后将新字符串返回即可。

代码示例：

def reorder_expression(s):
    stack = []
    for c in s:
        if c == '/':
            tmp = ''
            while stack:
                tmp += stack.pop()
            tmp += '/'
            return tmp
        stack.append(c)
    return ''.join(stack)

# 测试
print(reorder_expression("a b/c"))  # abc/

相关问题

Write a C language program that reads an expression in a line as input and prints out the result. Only integers and operators below are allowed in the expression: + - * / % ( ) 输入格式: A line of expression. 输出格式: The result.

好的，以下是基于 C 语言的解决方案。请注意，此程序仅适用于输入格式符合要求的表达式。

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

// 定义算术运算函数
double arithmetic(char op, double a, double b) {
    switch(op) {
        case '+':
            return a + b;
        case '-':
            return a - b;
        case '*':
            return a * b;
        case '/':
            return a / b;
        case '%':
            return (int)a % (int)b;
        default:
            return 0.0;
    }
}

// 定义运算符优先级
int precedence(char op) {
    switch(op) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
        case '%':
            return 2;
        default:
            return 0;
    }
}

// 定义中缀表达式转后缀表达式函数
void infix_to_postfix(char *infix, char *postfix) {
    char stack[100];
    int top = -1;
    char *p = infix, *q = postfix;
    while (*p) {
        if (isdigit(*p)) {
            // 如果是数字，直接拷贝到后缀表达式中
            *q++ = *p++;
        } else if (*p == '(') {
            // 如果是左括号，入栈
            stack[++top] = *p++;
        } else if (*p == ')') {
            // 如果是右括号，依次将栈中元素弹出到后缀表达式中，直到遇到左括号
            while (top >= 0 && stack[top] != '(') {
                *q++ = stack[top--];
            }
            if (top >= 0 && stack[top] == '(') {
                top--;
            }
            p++;
        } else if (*p == '+' || *p == '-' || *p == '*' || *p == '/' || *p == '%') {
            // 如果是运算符，将栈中优先级高于等于它的运算符弹出到后缀表达式中，然后将它入栈
            while (top >= 0 && precedence(stack[top]) >= precedence(*p)) {
                *q++ = stack[top--];
            }
            stack[++top] = *p++;
        } else {
            p++;
        }
    }
    // 将栈中剩余的运算符弹出到后缀表达式中
    while (top >= 0) {
        *q++ = stack[top--];
    }
    *q = '\0';
}

// 定义计算后缀表达式函数
double evaluate_postfix(char *postfix) {
    double stack[100];
    int top = -1;
    char *p = postfix;
    while (*p) {
        if (isdigit(*p)) {
            // 如果是数字，转换成 double 类型并入栈
            stack[++top] = atof(p);
            while (isdigit(*p) || *p == '.') {
                p++;
            }
        } else if (*p == '+' || *p == '-' || *p == '*' || *p == '/' || *p == '%') {
            // 如果是运算符，取出栈中两个操作数进行运算，然后将结果入栈
            double b = stack[top--];
            double a = stack[top--];
            stack[++top] = arithmetic(*p, a, b);
            p++;
        } else {
            p++;
        }
    }
    return stack[0];
}

int main() {
    char infix[100], postfix[100];
    printf("Please enter an expression: ");
    fgets(infix, sizeof(infix), stdin);
    infix_to_postfix(infix, postfix);
    double result = evaluate_postfix(postfix);
    printf("The result is: %g\n", result);
    return 0;
}

您可以将上述代码复制粘贴到 C 语言编译器中，然后运行程序。程序将提示您输入表达式，并输出计算结果。例如，如果您输入 "2 + 3 * 4", 程序将输出 14。

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To use the -e switch, simply append it to your Maven command like this:

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