表达式求值

1.题目简述

2.算法思路

3.代码

//
// Created by trmbh on 2023-09-11.
//

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

#define FALSE 0
#define TRUE 1
#define StackElementType char

typedef struct Node {
    StackElementType data;
    struct Node *next;
} LinkStackNode, *LinkStack;

/* 初始化栈区 */
int InitLinkStack(LinkStack *top) {
    *top = (LinkStack) malloc(sizeof(LinkStackNode));
    if (top == NULL) return FALSE
    (*top)->next = NULL;
    return TRUE
}

/* 入栈操作 */
int Push(LinkStack top, StackElementType x) {
    LinkStackNode *temp;
    temp = (LinkStackNode *) malloc(sizeof(LinkStackNode));
    temp->data = x;
    temp->next = top->next;
    top->next = temp;
    return TRUE;
}

/* 出栈操作 */
int Pop(LinkStack top, StackElementType *x) {
    if (top->next == NULL) return FALSE;
    LinkStackNode *temp;
    temp = top->next;
    top->next = temp->next;
    *x = temp->data;
    free(temp);
    return TRUE;
}

/* 获得栈首数据 */
int GetTop(LinkStack top, StackElementType *x) {
    if (top->next == NULL) return FALSE;
    *x = top->next->data;
    return TRUE;
}

/* 判断操作符 */
int IsOPTR(char c) {
    switch (c) {
        case '+':
        case '-':
        case '*':
        case '/':
        case '^':  // 添加幂运算操作符
        case '#':
            return TRUE;  // 是操作符
        default:
            return FALSE;  // 不是操作符
    }
}

/* 判断运算符优先级 */
char Compare(char c1, char c2) {
    switch (c1) {
        case '#':
            if (c2 == '#') return '=';
            else return '<';
        case '+':
        case '-':
            if (c2 == '#') return '>';
            else if (c2 == '+' || c2 == '-') return '=';
            else return '<';
        case '*':
        case '/':
            if (c2 == '#' || c2 == '+' || c2 == '-') return '>';
            else if (c2 == '*' || c2 == '/') return '=';
            else return '<';
        case '^':
            if (c2 == '^') return '=';
            else return '>';
        default:
            return '!';
    }
}

/* 判断是否为空 */
int isEmpty(LinkStack top) {
    if (top->data == '#') return FALSE;
    else return TRUE;
}

/* 表达式运算 */
int PerformOperation(int x, int y, char c) {
    switch (c) {
        case '^':
            return pow(x, y);
        case '*':
            return x * y;
        case '/':
            if (y != 0) return x / y;
            else return FALSE;
        case '+':
            return x + y;
        case '-':
            return x - y;
    }
}

/* 表达式求值 */

int EvaluateExpression(char *expression) {
    LinkStack operandStack, operatorStack;
    StackElementType currentChar, nextChar, topOperator;
    int operand1, operand2, result;
    char opChar1,opChar2;
    /* 初始化操作数栈和符号栈 */
    InitLinkStack(&operandStack);
    InitLinkStack(&operatorStack);

    /* 先向符号栈导入#，方便后续操作*/
    Push(operatorStack, '#');
    GetTop(operatorStack, &topOperator);

    while (*expression != '#' || topOperator != '#') {
        if (!IsOPTR(*expression)) {
            Push(operandStack, *expression);
        } else {
            switch (Compare(*expression, topOperator)) {
                case '<':
                case '=':
                    Pop(operandStack, &opChar1);
                    Pop(operandStack, &opChar2);
                    Pop(operatorStack, &topOperator);
                    operand1 = (int) (opChar1 - '0');
                    operand2 = (int) (opChar2 - '0');
                    result = PerformOperation(operand2, operand1, topOperator);
                    Push(operandStack, (char)(result + '0'));
                    if (*expression != '#') {
                        Push(operatorStack, *expression);
                    }
                    break;
                case '>':
                    Push(operatorStack, *expression);
                    break;
            }
        }
        if (*expression != '#') {
            expression++;
        }
        GetTop(operatorStack, &topOperator);
    }

    GetTop(operandStack, &currentChar);
    return (int)(currentChar - '0');
}



int main() {
    char str[100];
    printf("请输入表达式(以#结尾):");

    /* 第二个参数为最大字符数限制， 第三个参数为从标准输入流读入*/
    fgets(str, sizeof(str), stdin); // 使用fgets接收输入
    // 移除输入字符串中的换行符
    size_t len = strlen(str);
    if (len > 0 && str[len - 1] == '\n') {
        str[len - 1] = '\0';
    }
    printf("表达式计算结果:%d\n", EvaluateExpression(str));
    return 0;
}