evaluate postfix expression

后缀表达式求值是指使用逆波兰表达式来计算给定的数学表达式。在后缀表达式中，操作符位于操作数之后，这使得计算机能够轻松地计算表达式的结果。后缀表达式求值的具体过程是：遍历表达式中的每个元素，如果该元素是数字，则将其压入堆栈，如果该元素是操作符，则从堆栈中弹出两个数，用该操作符对它们做相应的计算（根据遇到的操作符不同进行不同的计算），并将结果入栈；重复步骤2至5，直到表达式最右边，最后运算得出的值即为表达式的结果。

相关问题

Write a program to evaluate a postfix expression. You only have to handle four kinds of operators: +, -, x, and /. Format of functions: ElementType EvalPostfix( char *expr ); where expr points to a string that stores the postfix expression. It is guaranteed that there is exactly one space between any two operators or operands. The function EvalPostfix is supposed to return the value of the expression. If it is not a legal postfix expression, EvalPostfix must return a special value Infinity which is defined by the judge program.

Sure, I can help you with that. Here's the implementation of the `EvalPostfix` function in C:

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

#define Infinity 1e9 // special value for invalid expression

typedef double ElementType; // type of operands

typedef struct Stack {
    int top; // index of the top element
    int capacity; // max capacity of the stack
    ElementType *array; // array to store elements
} Stack;

// function to create a new stack
Stack *CreateStack(int capacity) {
    Stack *stack = (Stack *) malloc(sizeof(Stack));
    stack->top = -1;
    stack->capacity = capacity;
    stack->array = (ElementType *) malloc(stack->capacity * sizeof(ElementType));
    return stack;
}

// function to check if the stack is empty
bool IsEmpty(Stack *stack) {
    return stack->top == -1;
}

// function to check if the stack is full
bool IsFull(Stack *stack) {
    return stack->top == stack->capacity - 1;
}

// function to push an element onto the stack
void Push(Stack *stack, ElementType element) {
    if (IsFull(stack)) {
        printf("Error: Stack is full\n");
        exit(EXIT_FAILURE);
    }
    stack->array[++stack->top] = element;
}

// function to pop an element from the stack
ElementType Pop(Stack *stack) {
    if (IsEmpty(stack)) {
        printf("Error: Stack is empty\n");
        exit(EXIT_FAILURE);
    }
    return stack->array[stack->top--];
}

// function to evaluate a postfix expression
ElementType EvalPostfix(char *expr) {
    Stack *stack = CreateStack(strlen(expr));
    char *token = strtok(expr, " "); // get the first token
    while (token != NULL) {
        if (isdigit(token[0])) { // if token is an operand
            Push(stack, atof(token)); // push it onto the stack
        } else { // if token is an operator
            // pop the top two elements from the stack
            ElementType operand2 = Pop(stack);
            ElementType operand1 = Pop(stack);
            // perform the operation and push the result onto the stack
            switch (token[0]) {
                case '+':
                    Push(stack, operand1 + operand2);
                    break;
                case '-':
                    Push(stack, operand1 - operand2);
                    break;
                case 'x':
                    Push(stack, operand1 * operand2);
                    break;
                case '/':
                    if (operand2 == 0) { // division by zero
                        return Infinity;
                    }
                    Push(stack, operand1 / operand2);
                    break;
                default:
                    return Infinity; // invalid operator
            }
        }
        token = strtok(NULL, " "); // get the next token
    }
    if (stack->top != 0) { // invalid expression
        return Infinity;
    }
    return Pop(stack); // return the final result
}

int main() {
    char expr[] = "5 3 2 x - 4 +";
    ElementType result = EvalPostfix(expr);
    if (result == Infinity) {
        printf("Error: Invalid expression\n");
    } else {
        printf("Result = %g\n", result);
    }
    return 0;
}

In this implementation, we use a stack to store the operands and perform the operations on them. We tokenize the postfix expression using the `strtok` function and process each token one by one. If the token is an operand, we convert it to a `double` and push it onto the stack. If the token is an operator, we pop the top two elements from the stack, perform the operation, and push the result back onto the stack. If the expression is invalid, we return the special value `Infinity`.

这是上题的代码：def infix_to_postfix(expression): precedence = {'!': 3, '&': 2, '|': 1, '(': 0} op_stack = [] postfix_list = [] token_list = expression.split() for token in token_list: if token.isalnum(): postfix_list.append(token) elif token == '(': op_stack.append(token) elif token == ')': top_token = op_stack.pop() while top_token != '(': postfix_list.append(top_token) top_token = op_stack.pop() else: # operator while op_stack and precedence[op_stack[-1]] >= precedence[token]: postfix_list.append(op_stack.pop()) op_stack.append(token) while op_stack: postfix_list.append(op_stack.pop()) return ' '.join(postfix_list) class Node: def __init__(self, value): self.value = value self.left_child = None self.right_child = None def build_expression_tree(postfix_expr): operator_stack = [] token_list = postfix_expr.split() for token in token_list: if token.isalnum(): node = Node(token) operator_stack.append(node) else: right_node = operator_stack.pop() left_node = operator_stack.pop() node = Node(token) node.left_child = left_node node.right_child = right_node operator_stack.append(node) return operator_stack.pop() def evaluate_expression_tree(node, variable_values): if node.value.isalnum(): return variable_values[node.value] else: left_value = evaluate_expression_tree(node.left_child, variable_values) right_value = evaluate_expression_tree(node.right_child, variable_values) if node.value == '!': return not left_value elif node.value == '&': return left_value and right_value elif node.value == '|': return left_value or right_value expression = "!a & (b | c)" postfix_expression = infix_to_postfix(expression) expression_tree = build_expression_tree(postfix_expression) variable_values = {'a': True, 'b': False, 'c': True} result = evaluate_expression_tree(expression_tree, variable_values) print(result)

好的，感谢提供代码。以下是针对你的问题的解答：

1. 打印二叉树构造过程

在 build_expression_tree 函数中，operator_stack 列表存储了构建二叉树的过程。我们可以在函数中添加打印语句来查看二叉树构建过程。

代码如下：

def build_expression_tree(postfix_expr):
    operator_stack = []
    token_list = postfix_expr.split()

    for token in token_list:
        if token.isalnum():
            node = Node(token)
            operator_stack.append(node)
        else:
            right_node = operator_stack.pop()
            left_node = operator_stack.pop()
            node = Node(token)
            node.left_child = left_node
            node.right_child = right_node
            operator_stack.append(node)
        
        # 打印二叉树构建过程
        print([i.value for i in operator_stack])

    return operator_stack.pop()

在运行 build_expression_tree 函数时，将会打印出二叉树构建过程。

2. 打印后缀表达式和二叉树后序遍历序列

在函数 infix_to_postfix 中，postfix_list 列表存储了后缀表达式。我们可以在函数返回后缀表达式的同时，打印出后缀表达式。

代码如下：

def infix_to_postfix(expression):
    precedence = {'!': 3, '&': 2, '|': 1, '(': 0}
    op_stack = []
    postfix_list = []
    token_list = expression.split()

    for token in token_list:
        if token.isalnum():
            postfix_list.append(token)
        elif token == '(':
            op_stack.append(token)
        elif token == ')':
            top_token = op_stack.pop()
            while top_token != '(':
                postfix_list.append(top_token)
                top_token = op_stack.pop()
        else:
            # operator
            while op_stack and precedence[op_stack[-1]] >= precedence[token]:
                postfix_list.append(op_stack.pop())
            op_stack.append(token)

    while op_stack:
        postfix_list.append(op_stack.pop())

    # 打印后缀表达式
    print(f"后缀表达式：{' '.join(postfix_list)}")

    return ' '.join(postfix_list)


def build_expression_tree(postfix_expr):
    operator_stack = []
    token_list = postfix_expr.split()

    for token in token_list:
        if token.isalnum():
            node = Node(token)
            operator_stack.append(node)
        else:
            right_node = operator_stack.pop()
            left_node = operator_stack.pop()
            node = Node(token)
            node.left_child = left_node
            node.right_child = right_node
            operator_stack.append(node)
        
        # 打印二叉树构建过程
        print([i.value for i in operator_stack])

    return operator_stack.pop()

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)

# 打印二叉树后序遍历序列
print("二叉树后序遍历序列：", end="")
def postorder(node):
    if node:
        postorder(node.left_child)
        postorder(node.right_child)
        print(node.value, end=" ")
postorder(expression_tree)
print()

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，最后打印出二叉树的后序遍历序列。

3. 打印二叉树评估过程

在 evaluate_expression_tree 函数中，我们可以添加打印语句来查看二叉树评估的过程。

代码如下：

def evaluate_expression_tree(node, variable_values):
    if node.value.isalnum():
        return variable_values[node.value]
    else:
        left_value = evaluate_expression_tree(node.left_child, variable_values)
        right_value = evaluate_expression_tree(node.right_child, variable_values)
        if node.value == '!':
            result = not left_value
        elif node.value == '&':
            result = left_value and right_value
        elif node.value == '|':
            result = left_value or right_value

        # 打印二叉树评估过程
        print(f"节点 {node.value} 的值为 {result}")

        return result

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)
variable_values = {'a': True, 'b': False, 'c': True}
result = evaluate_expression_tree(expression_tree, variable_values)
print(result)

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，接着会打印出二叉树评估的过程。

4. 显示公式真值表

可以通过遍历变量的所有取值，来计算表达式的真值表。

代码如下：

expression = "!a & (b | c)"
postfix_expression = infix_to_postfix(expression)
expression_tree = build_expression_tree(postfix_expression)

# 遍历变量的所有可能取值
variable_values = {'a': True, 'b': False, 'c': True}
variable_names = list(variable_values.keys())
print(" ".join(variable_names), expression)

for i in range(2**len(variable_names)):
    values = [bool(int(x)) for x in bin(i)[2:].zfill(len(variable_names))]
    variable_values = dict(zip(variable_names, values))
    result = evaluate_expression_tree(expression_tree, variable_values)
    print(" ".join([str(int(x)) for x in values]), int(result))
    

在运行程序时，将会先打印出后缀表达式，然后是构建二叉树的过程，接着会打印出真值表。