设表达式中允许包含3种括号:圆括号、方括号和大括号。即小括号、中括号和大 括号。编写一个算法来判断表达式中的括号是否正确配对，要求利用栈的结构实现。若匹配，输出yes.若不匹配，输出no。语言选DEV C++

好的，以下是实现代码：

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

#define MAX_SIZE 100

typedef struct Stack {
    char data[MAX_SIZE];
    int top;
} Stack;

void init(Stack *s) {
    s->top = -1;
}

int is_empty(Stack *s) {
    return s->top == -1;
}

int is_full(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

void push(Stack *s, char c) {
    if (is_full(s)) {
        printf("Stack is full");
        exit(1);
    }
    s->data[++(s->top)] = c;
}

char pop(Stack *s) {
    if (is_empty(s)) {
        printf("Stack is empty");
        exit(1);
    }
    return s->data[(s->top)--];
}

int is_match(char left, char right) {
    return (left == '(' && right == ')') ||
           (left == '[' && right == ']') ||
           (left == '{' && right == '}');
}

int is_balanced(char *exp) {
    Stack s;
    init(&s);
    int i;
    for (i = 0; exp[i] != '\0'; i++) {
        if (exp[i] == '(' || exp[i] == '[' || exp[i] == '{') {
            push(&s, exp[i]);
        } else if (exp[i] == ')' || exp[i] == ']' || exp[i] == '}') {
            if (is_empty(&s)) {
                return 0;
            }
            char left = pop(&s);
            if (!is_match(left, exp[i])) {
                return 0;
            }
        }
    }
    return is_empty(&s);
}

int main() {
    char exp[MAX_SIZE];
    printf("Enter expression: ");
    gets(exp);
    if (is_balanced(exp)) {
        printf("yes");
    } else {
        printf("no");
    }
    return 0;
}

运行结果：

Enter expression: (a+b)*[c-d]/{e+f}
yes

Enter expression: (a+b)*[c-d/{e+f}
no