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c++ - 以相反的顺序显示内容

问题描述

下面的代码是由一位溢出者向我建议的。所以学分不是我的。我试图绕过这段代码并尝试以相反的顺序打印出元素。到目前为止，这些元素都是从起始词 dog 开始打印出来的。但目标是以另一种方式打印它。从猫开始。因此，基本上，代码可以根据每个人的祖先来回溯单词。例如，在这种情况下，我们从 cag 中得到 cat，它是祖先，而 cag 的祖先是 cog。依此类推，直到我们从狗开始

#include <iostream>
#include <string>
#include <unordered_set>
#include <stack>
#include <vector>

using namespace std;

int main() {
    vector<string> dictionary;
    vector<pair<string, int>> words; //stores (word, predecessor)
    string startWord = "dog";
    string endWord = "cat";

    unordered_set<string> seenWords;

    dictionary.push_back("dog");
    dictionary.push_back("bog");
    dictionary.push_back("cog");
    dictionary.push_back("fog");
    dictionary.push_back("cat");
    dictionary.push_back("bag");
    dictionary.push_back("beg");
    dictionary.push_back("bet");
    dictionary.push_back("bat");


    words.emplace_back(startWord, -1);
    seenWords.insert(startWord);

    bool found = false;

    //Try all new words as reference words
    for(int i = 0; i < words.size() && !found; ++i) {       
        //we look for words that we can generate from words[i]
        cout << i << " " << words[i].first << ":   ";

        //try all the words from the dictionary
        for (int j = 0; j < dictionary.size(); j++) {
            string& candidate = dictionary[j];
            //check if candidate can be generated from reference

            //count the different characters
            int differentCharacters = 0;
            for (int pos = 0; pos < words[i].first.size(); ++pos)
            {
                if (candidate[pos] != words[i].first[pos])
                    ++differentCharacters;
            }
            if (differentCharacters == 1 && seenWords.find(candidate) == seenWords.end()) {
                //yes, we can generate this candidate from word[i] and we haven't seen the word before
                cout << "(" << words.size() << ")" << candidate << " ";                         

                words.emplace_back(candidate, i);
                seenWords.insert(candidate);

                if (candidate == endWord) {
                    found = true;
                    cout << "Found endword";
                    break;
                }
            }           
        }
        cout << endl;
    }

    if (found) {
        //traverse the word path from the end word back to the start word
        int i = words.size() - 1;
        stack<string> wordPath;
        while (i != -1) {
            //push the current word onto a stack
            wordPath.push(words[i].first);
            //go to the previous word
            i = words[i].second;
        }

        //now retrieve the words from the stack and print them in reverse order
        cout << "Word path:" << endl;
        while (!wordPath.empty()) {
            cout << wordPath.top() << " ";
            wordPath.pop();
        }
        cout << endl;
    }

    return EXIT_SUCCESS;
}

标签： c++

解决方案

它实际上非常简单！stack与其使用 a推送然后弹出您的“找到”字符串路径，不如使用avector和push_back字符串；然后您可以按任意顺序打印值！在这段代码中，我已经从你所拥有的切换到“其他”顺序：

if (found) {
    //traverse the word path from the end word back to the start word
    int i = words.size() - 1;
/// stack<string> wordPath;
    vector<string> wordPath;
    while (i != -1) {
        // push the current word into a vector ...
///     wordPath.push(words[i].first);
        wordPath.push_back(words[i].first);
        //go to the previous word
        i = words[i].second;
    }
    // now retrieve the words from the vector and print them ...
    cout << "Word path:" << endl;
/// while (!wordPath.empty()) {
///     cout << wordPath.top() << " ";
///     wordPath.pop();
/// }
    ///
    for (size_t w = 0; w < wordPath.size(); ++w) {
        string text = wordPath[w];
        size_t index = 0;
        for (index = 0; index < dictionary.size(); ++index) {
            if (text == dictionary[index]) break;
        }
        cout << text << "[" << index << "] ";
    }
    ///
    cout << endl;
}

你甚至可以在这里做出选择！要以原始（='reverse'）顺序打印，只需更改for循环：

    for (size_t w = wordPath.size() - 1; w <= 0 ; --w) {
        cout << wordPath[w] << " ";
    }

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