Question

我正在尝试用蛮力方法解决博客问题。输入是：

表示为字符串的网格矩阵，例如：'pesa' 这个词，也表示为字符串'asa'。

我正在编写一个函数来检查该单词是否是矩阵中的合法单词。

bool Boogle::contains(std::string grid, std::string word) const
{
    bool* isvisited=new bool[grid.length()];
    for (unsigned int i=0; i<grid.length(); i++)
    {
        *(isvisited+i)=false;
    }

    for (unsigned int i=0; i<grid.length(); i++)
    {
        // Recursive approach
        if (grid[i]==word[0])
            if (checkqueue(grid, word, isvisited, i, 0))
                return true;    
    }
    return false;
}

bool Boogle::checkqueue(const string &grid, const string &word, bool* const &isvisited, unsigned int grid_index, unsigned int count) const
{   
    int matsize=int(sqrt(grid.length()));
    cout<<"\nCurrently at the index "<<grid_index<<"\n";
    isvisited[grid_index]=true;
    for (unsigned int i=0; i<grid.length(); i++)
    {
        cout <<isvisited[i]<<" ";
    }
    cout<<"\n";

    if (count==word.length()-1)
    {
        cout << " reach the end of word\n";
        return true;
    }
    else
    {
        count ++;
        cout << "Recursive call on WORD: "<<word<<"  " <<count<<" "<<word[count]<<"\n";

        // non diagonal
        if ((grid_index<grid.length()) && (isvisited[grid_index+1]==false) && (grid[grid_index+1]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index+1, count);

        else if ((grid_index>0)&& (isvisited[grid_index-1]==false)  && (grid[grid_index-1]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index-1, count);

        else if (((grid_index+matsize)<grid.length())&& (isvisited[grid_index+matsize]==false)  && (grid[grid_index+matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index+1, count);

        else if (((grid_index-matsize)<grid.length())&& (isvisited[grid_index-matsize]==false)  && (grid[grid_index-matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index+1, count);

        // diagonal
        else if ((grid_index-1-matsize>0)&& (isvisited[grid_index-1-matsize]==false)  && (grid[grid_index-1-matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index-1-matsize, count);

        else if ((grid_index+1-matsize>0) && (isvisited[grid_index+1-matsize]==false) && (grid[grid_index+1-matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index+1-matsize, count);

        else if ((grid_index+1+matsize<grid.length())&& (isvisited[grid_index+1+matsize]==false)  && (grid[grid_index+1+matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index+1+matsize, count);

        else if ((grid_index-1+matsize<grid.length())&& (isvisited[grid_index-1+matsize]==false)  && (grid[grid_index-1+matsize]==word[count]))
            return checkqueue(grid, word, isvisited, grid_index-1+matsize, count);
        else
        {
            // cout<<"No possible neighbor\n";
            return false;
        }

    }
}

如果我运行boogle.contains（“pesa”，“as”），它的效果很好。但如果这是一个非法的词，如“asa”，它会返回分段错误。它来自哪里？

./Boogle.exe 
.
Currently at the index 3
0 0 0 1 
Recursive call on WORD: asa  1 s

Currently at the index 2
0 0 1 1 
Recursive call on WORD: asa  2 a
Segmentation fault: 11

P / S：当单词有效时，这是正确的运行（boogle.contains（“pesa”，“esp”））

Currently at the index 1
0 1 0 0 
Recursive call on WORD: esp  1 s

Currently at the index 2
0 1 1 0 
Recursive call on WORD: esp  2 p

Currently at the index 3
0 1 1 1 
 reach the end of word



OK (1 tests)

Answer 1

您正在混合有符号和无符号算术。 matsize为int，grid_index为unsigned int。您与>0的比较无效，因为结果总是无符号的，因此从不消极。

顺便说一下，修复签名/未签名的不匹配后，您可能需要>=0。

Answer 2

如果您尝试在条件列表前显示cout << grid_index-1-matsize << endl;。您将看到由于grid_index未签名而发生缓冲区下溢。

出于这个原因，条件(grid_index-matsize)<grid.length()为真，因为grid_index已超过grid.length()。

您必须将函数checkqueue的签名更改为

bool Boogle::checkqueue(const string &grid, const string &word, bool* const &isvisited, int grid_index, unsigned int count) const

博客中的分段错误

2 个答案: