Question

所以我正在为一个项目编写霍夫曼编码。但是，我的代码不起作用。当我在visual studio上运行它时，它并没有给我一个错误。我试图做的是读取一个文件并将所有文件放入一个字符串中。并获取该字符串中每个字符的频率。但我认为当文件有点大时，似乎我的代码在无限循环中运行。任何人都可以向我解释一下吗？顺便说一下，我有一个排序函数，我用它按频率对节点*的矢量进行排序。

ifstream infile;
infile.open(filename);
string q;
string line;
while (getline(infile, line))
{
    q += line;
}
char y;
int count = 0;
int check = 0;
for (int i = 0; i < q.size(); i++) //if the string gets big, it seems to become an infinite loop in here
{
    y = q[i];
    for (int x = i - 1; x > 0; x--)  //make sure not counting the same char
    {
        if (y == q[x])
        {
            check++;
        }
    }
    if (check == 0)
    {
        for (int i = 0; i < q.size(); i++)
        {
            if (q[i] == y)
            {
                count++;
            }
        }
        node*x = new node;
        x->char1 = y;   //my node have char 
        x->freq = count; //my node has frequency
        list1.push_back(x);
    }
    count = 0;
    check = 0;
}
sort(list1.begin(), list1.end(), sorter);  //sort them from small to big
while (list1.size() > 1)
{
    node*left = list1[0];
    node*right = list1[1];
    list1.erase(list1.begin(), list1.begin() + 2);
    double sum = left->freq + right->freq;
    node* x = new node;
    x->freq = sum;
    x->left = left;
    x->right = right;
    list1.push_back(x);
    sort(list1.begin(), list1.end(), sorter);
}
list1.clear();
return true;

以下是我的排序功能

static struct {
bool operator()(NodeInterface* a, NodeInterface* b) {
    if (a->getFrequency() == b->getFrequency()) {//if the frequencies are even,
        if (b->getCharacter() == '\0') return false;
        if (a->getCharacter() != '\0') {
            return (int)a->getCharacter() < (int)b->getCharacter();

        }
        return false;
    }
    return a->getFrequency() < b->getFrequency();
}

}分拣机;

Answer 1

我看到两个主要问题。

在for循环中你有一个for循环初始化和使用int i

更改内循环的变量名称。

for (int i = 0; i < q.size(); i++) //if the string gets big, it seems to become an infinite loop in here
.
.
    if (check == 0)
    {
        for (int i = 0; i < q.size(); i++) //Change this to int j for example
        {
        .
        .

和Sorter结构。我会把它改写成这个。

static struct {
    bool operator()(NodeInterface* a, NodeInterface* b) {
        if (a->getFrequency() == b->getFrequency()) {//if the frequencies are even,
            if (b->getCharacter() == '\0') return false;
            if (a->getCharacter() == '\0') return true;
            return (int)a->getCharacter() < (int)b->getCharacter();
        }
        return a->getFrequency() < b->getFrequency();
    }
} sorter;

针对你的for循环的一些建议：

for (int i = 0; i < q.size(); i++) //if the string gets big, it seems to become an infinite loop in here
{
    y = q[i];
    //You can avoid this entire loop by using a structure like map
    for (int x = i - 1; x > 0; x--)  //make sure not counting the same char
    {
        if (y == q[x])
        {
        check++;
        //break; //if you use a loop, break it once you find the character.
        }
    }
    if (check == 0)
    {
        for (int j = 0; j < q.size(); j++)//Renamed variable + you can start this loop from j = i as you know there is no occurrence of y before that.
        {
            if (q[i] == y)
            {
                count++;
            }
        }
        node*x = new node;
        x->char1 = y;   //my node have char 
        x->freq = count; //my node has frequency
        list1.push_back(x);
    }
    count = 0;
    check = 0;
}

霍夫曼编码c ++

1 个答案: