Question

下面的代码是由同一个溢出者建议给我的。因此，积分不是我的。我试图绕过这段代码，然后尝试以相反的顺序打印出元素。到目前为止，这些元素都是从狗的起始单词开始打印出来的。但目标是将其打印为其他方式。从猫开始。因此，从本质上讲，代码的工作方式是根据每个单词的祖先回溯单词。例如，在这种情况下，我们从作为祖先的cag那里得到了cat，而cag的祖先是cog。依此类推，直到我们开始养狗

#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;
}

Answer 1

实际上非常简单！无需使用{ "_id" : ObjectId("5d9f074f5833c8cd1f685e05"), "tags" : [ { "Domain" : "http://www.google.com", "rank" : 1 }, { "Domain" : "https://www.stackoverflow.com/", "rank" : 2 } ] }来推送然后弹出“找到的”字符串路径，而只需使用stack和vector字符串即可；然后您就可以按两种顺序打印出值！在这段代码中，我已从原来的顺序切换到“其他”顺序：

push_back

您甚至可以在这里做出选择！要按原始（=“反向”）顺序打印，只需更改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; }循环：

for

Answer 2

您可以简单地使用.rbegin(), .rend()和reverse_iterator后退dictionary，并在到达"cat"时使用标志开始打印。参见std::vector，例如

#include <iostream>
#include <vector>
#include <string>

int main () {

    bool prn = false;
    std::vector<std::string> dictionary;

    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");

    /* output in reverse order beginning with cat */
    for (auto it = dictionary.rbegin(); it != dictionary.rend(); it++) {
        if (*it == "cat")
            prn = true;
        if (prn)
            std::cout << *it << '\n';
    }
}

使用/输出示例

如果我理解并且您只是想以"cat"开始以相反的顺序打印，那么以下内容应与您尝试执行的操作相符：

$ ./bin/reverse_cats
cat
fog
cog
bog
dog

输出关联矢量索引

如果要与字符串一起输出矢量索引，则可以使用dictionary.rend() - it - 1获取从零开始的索引，例如

    /* output in reverse order beginning with cat */
    for (auto it = dictionary.rbegin(); it != dictionary.rend(); it++) {
        if (*it == "cat")
            prn = true;
        if (prn)
            std::cout << dictionary.rend() - it - 1 << " " << *it << '\n';
    }

使用/输出示例

$ ./bin/reverse_cats
4 cat
3 fog
2 cog
1 bog
0 dog

以相反的顺序显示内容

2 个答案: