数据结构用栈实现表达式求值C代码

以下是使用栈实现表达式求值的C代码：

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

#define MAX_SIZE 100

// 定义栈结构
typedef struct {
    int data[MAX_SIZE];
    int top;
} Stack;

// 初始化栈
void initStack(Stack *s) {
    s->top = -1;
}

// 判断栈是否为空
int isEmpty(Stack *s) {
    return s->top == -1;
}

// 判断栈是否已满
int isFull(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

// 入栈
void push(Stack *s, int value) {
    if (isFull(s)) {
        printf("Stack is full.\n");
        exit(1);
    }
    s->data[++s->top] = value;
}

// 出栈
int pop(Stack *s) {
    if (isEmpty(s)) {
        printf("Stack is empty.\n");        exit(1);
    }
    return s->data[s->top--];
}

// 获取栈顶元素
int peek(Stack *s) {
    if (isEmpty(s)) {
        printf("Stack is empty.\n");
        exit(1);
    }
    return s->data[s->top];
}

// 判断运算符的优先级
int getPriority(char op) {
    switch (op) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        default:
            return 0;
    }
}

// 执行运算
int calculate(int num1, int num2, char op) {
    switch (op) {
        case '+':
            return num1 + num2;
        case '-':
            return num1 - num2;
        case '*':
            return num1 * num2;
        case '/':
            return num1 / num2;
        default:
            return 0;
    }
}

// 表达式求值
int evaluateExpression(char *expression) {
    Stack operandStack; // 操作数栈
    Stack operatorStack; // 运算符栈
    initStack(&operandStack);
    initStack(&operatorStack);

    int i = 0;
    while (expression[i] != '#') {
        if (expression[i] >= '0' && expression[i] <= '9') {
            // 如果是数字，直接入操作数栈
            push(&operandStack, expression[i] - '0');        } else if (expression[i] == '+' || expression[i] == '-' || expression[i] == '*' || expression[i] == '/') {
            // 如果是运算符
            while (!isEmpty(&operatorStack) && getPriority(expression[i]) <= getPriority(peek(&operatorStack))) {
                // 当前运算符的优先级小于等于栈顶运算符的优先级，进行运算
                int num2 = pop(&operandStack);
                int num1 = pop(&operandStack);
                char op = pop(&operatorStack);
                int result = calculate(num1, num2, op);
                push(&operandStack, result);
            }
            // 当前运算符入运算符栈
            push(&operatorStack, expression[i]);
        }
        i++;
    }

    // 表达式遍历完毕，进行最后的运算
    while (!isEmpty(&operatorStack)) {
        int num2 = pop(&operandStack);
        int num1 = pop(&operandStack);
        char op = pop(&operatorStack);
        int result = calculate(num1, num2, op);
        push(&operandStack, result);
    }

    // 返回最终结果
    return pop(&operandStack);
}

int main() {
    char expression[MAX_SIZE];
    printf("请输入表达式（以#结尾）：");
    scanf("%s", expression);
    int result = evaluateExpression(expression);
    printf("表达式的值为：%d\n", result);
    return 0;
}