具有过度约束类的噩梦表达式树

Question

我无意中让我的学生过度约束用于解决以下问题的共享类。我意识到这可能是该网站的用户可能喜欢的一个问题。

第一个团队/函数 getNodes 采用一个表示使用有符号整数和四个操作 +、-、* 和 / 的前缀表达式的字符串，并使用 Node 类生成相应的以空结尾的标记链表，其中标记通过“右”指针。

第二个团队/函数 getTree 采用类似的字符串，将其传递给 getNodes，并将结果节点重新链接为表达式树。

第三个团队/函数，evaluate，采用类似的字符串，将其传递给 getTree，并评估结果表达式树以形成答案。

过度约束的 exptree.h 如下。这个问题必须通过只编写上面定义的三个函数来解决，不需要额外的函数。

#ifndef EXPTREE_H_
#define EXPTREE_H_

using namespace std;

enum Ops{ADD, SUB, MUL, DIV, NUM};

class Node {
   private:
        int num;
        Ops op;
        Node *left, *right;

    public:
        friend Node *getNodes(string d);
        friend Node *getTree(string d);
        friend int evaluate (string);
    };

int evaluate(string d);
Node *getNodes(string d);
Node *getTree(string d);
#endif

唯一可以使用的库是这些。

#include <iostream>
#include <vector>
#include <string>
#include "exptree.h" 

对于那些担心我的学生的人，我今天将指出，只需几个更合适的函数就可以轻松解决这个问题。我知道表达式树可以编码有理数而不仅仅是整数。今天我也会指出这一点。

这是我根据他们的规格给他们的驱动程序。

#include <iostream>
#include <string>
#include "exptree.h"
using namespace std;
void test(string s, int target) {
    int result = evaluate(s);
    if (result == target)
        cout << s << " correctly evaluates to " << target << endl;
    else
        cout << s << "(" << result 
             << ") incorrectly evaluates to " << target << endl;
}
int main() {
    test("42", 42);
    test("* - / 4 2 1 42", 42);
    test("* - / -4 +2 -1 2", -2);
    test("* - / -4 +2 -1 2            ", -2);
    test("* 9 6", 54);
    return 0;
}

你能以尽可能优雅的方式编写这三个函数来解决这个噩梦般的问题吗？

Answer 1

在这些约束下编写 getNodes 和 getTree 函数将非常简单，因此我直接跳到有趣的部分。您自然会递归地计算表达式树，但这不是这里的一个选项，因为 eval 函数只接受一个字符串。当然，您可以将剩余的树重新字符串化为前缀表达式，并对其递归调用 eval，但这只是愚蠢的。

首先，我将表达式树转换为后缀表达式，使用显式堆栈作为穷人的递归。然后我使用标准操作数堆栈对其进行评估。

#include <iostream>
#include <vector>
#include <string>
using namespace std;
#include "exptree.h" 

int evaluate(string d){
    Node* tree = getTree(d);
   //convert tree to postfix for simpler evaluation
    vector<Node*> node_stack;
    node_stack.push_back(tree);
    Node postfix_head;
    Node* postfix_tail = &postfix_head;
    while(node_stack.size() > 0){
        Node* place = node_stack.back();
        if(place->left == 0){
             if(place->right == 0){
                 postfix_tail->right = place;
                 node_stack.pop_back();
             } else {
                 node_stack.push_back(place->right);
                 place->right = 0;
             }
        } else {
            node_stack.push_back(place->left);
            place->left = 0;
        }
    }
   //evaluate postfix
    Node* place = postfix_head.right;
    vector<int> stack;
    while(place != 0){
        if(place->op != NUM){
            int operand_a, operand_b;
            operand_b = stack.back();
            stack.pop_back();
            operand_a = stack.back();
            stack.pop_back();
            switch(place->op){
                case ADD:
                    stack.push_back(operand_a + operand_b);
                    break;
                case SUB:
                    stack.push_back(operand_a - operand_b);
                    break;
                case MUL:
                    stack.push_back(operand_a * operand_b);
                    break;
                case DIV:
                    stack.push_back(operand_a / operand_b);
                    break;
            }
        } else {
            stack.push_back(place->num);
        }
        place = place->right;
    }
    return stack.back();
}

Answer 2

我认为“没有附加功能”这个要求太苛刻了。实现 getTree 的最简单方法可能是递归，并且需要定义一个附加函数。

Node* relink(Node* start) // builds a tree; returns the following node
{
    if (start->op == NUM)
    {
        Node* result = start->right;
        start->left = start->right = NULL;
        return result;
    }
    else
    {
        start->left = start->right;
        start->right = relink(start->left);
        return relink(start->right);
    }
}

Node* getTree(string d)
{
    Node* head = getNodes(d);
    relink(head);
    return head;
}

我可以通过使用显式堆栈（由 std::vector 实现）来实现递归，但这很丑陋且晦涩（除非您希望学生完全练习这一点）。

Answer 3

就其价值而言，这是我在发布问题之前编写的解决方案

#include <iostream>
#include <vector>
#include "exptree.h"
using namespace std;

Node *getNodes(string s) {
    const int MAXINT =(int)(((unsigned int)-1) >> 1), MININT = -MAXINT -1;
    Node *list;
    int sign, num;

    s += " ";                                                   // this simplifies a lot of logic, allows trailing white space to always close off an integer
    list = (Node *) (num = sign = 0);
    for (int i=0; i<s.size(); ++i) {
        char c = s[i];                                          // more efficient and cleaner reference to the current character under scrutiny
        if (isdigit(c)) {
            if (sign == 0) sign = 1;                            // if sign not set, then set it. A blank with a sign==0 now signifies a blank that can be skipped
            num = 10*num + c - '0';
        } else if (((c=='+') || (c=='-')) && isdigit(s[i+1])) { // another advantage of adding blank to string above so don't need a special case
            sign = (c=='+') ? 1 : -1;
        } else if ( !isspace(c) &&  (c != '+')  && (c != '-')  && (c != '*')  && (c != '/')) {
            cout << "unexpected character " << c << endl;
            exit(1);
        } else if  (!isspace(c) || (sign != 0)) {                                                       // have enough info to create next Node
            list->left = (list == 0) ? (list = new Node) : (list->left->right = new Node);              // make sure left pointer of first Node points to last Node
            list->left->right = 0;                                                                      // make sure list is still null terminated
            list->left->op = (c=='+' ? ADD : (c=='-' ? SUB : (c=='*' ? MUL : (c=='/' ? DIV : NUM))));   // choose right enumerated type
            list->left->num = (list->left->op==NUM) ? sign*num : MININT;                                // if interior node mark number for evaluate function
            num = sign = 0;                                                                             // prepare for next Node
        }
    }
    return list;
}

Node *getTree(string s) {
    Node *nodes = getNodes(s), *tree=0, *root, *node;
    vector<Node *> stack;

    if (nodes == 0) return tree;
    root = tree = nodes;
    nodes = nodes->right;
    for (node=nodes; node != 0; node=nodes) {
        nodes = nodes->right;
        if (root->op != NUM) {              // push interior operator Node on stack til time to point to its right tree
            stack.push_back(root);
            root = (root->left = node);     // set interior operator Node's left tree and prepare to process that left tree
        } else {                            
            root->left = root->right = 0;   // got a leaf number Node so finish it off
            if (stack.size() == 0) break;
            root = stack.back();            // now pop operator Node off the stack
            stack.pop_back();
            root = (root->right = node);    // set its left tree and prepare to process that left tree
        }
    }
    if ((stack.size() != 0) || (nodes != 0)) {
        cout << "prefix expression has missing or extra terms" << endl;
        exit(1);
    }
    return tree;
}

int evaluate(string s) {
    // MININT is reserved value signifying operator waiting for a left side value, low inpact since at edge of representable integers
    const int MAXINT =(int)(((unsigned int)-1) >> 1), MININT = -MAXINT -1;
    Node *tree = getTree(s);
    vector<Node *> stack;
    int v = 0;                              // this is value of a leaf node (a number) or the result of evaluating an interior node
    if (tree == 0) return v;
    do {
        v = tree->num;
        if (tree->op != NUM) {
            stack.push_back(tree);
            tree = tree->left;              // prepare to process the left subtree      
        } else while (stack.size() != 0) {  // this while loop zooms us up the right side as far as we can go (till we come up left side or are done)
            delete tree;                    // done with leaf node or an interior node we just finished evaluating
            tree = stack.back();            // get last interior node from stack
            if (tree->num == MININT) {      // means returning up left side of node, so save result for later
                tree->num = v;
                tree = tree->right;         // prepare to evaluate the right subtree
                break;                      // leave the "else while" for the outer "do while" which handles evaluating an expression tree
            } else {                        // coming up right side of an interior node (time to calculate)
                stack.pop_back();           // all done with interior node
                v = tree->op==ADD ? tree->num+v : (tree->op==SUB ? tree->num-v : (tree->op==MUL ? tree->num*v : tree->num/v)) ;
            }
        }
    } while (stack.size() != 0);
    return v;
}