我创建了一个数据包结构。我读取一个文件的文本并将每个单词插入一个节点,然后如果有相同的字符串则增加计数。但我的问题是,我只想输出一串相同的字符串和它被使用的次数。但是每当我使用我的删除功能时,它都会删除我文件中的所有内容,如果我不使用它,我会得到如下所示的输出。我不知道我做错了什么,有没有办法不输出重复的字符串?
ofstream output;
struct BagNode
{
string dataValue;
string dataCopy;
int dataCountCopy;
int dataCount;
BagNode * next;
};
class Bag{
private:
BagNode * head;
public:
Bag()
{
head = NULL;
}
void insert(string v)
{
if(head == NULL){ //empty list
head = new BagNode;
removePunct(v);
head->dataValue = v;
transform(v.begin(), v.end(), v.begin(), ::tolower);
head->dataCopy = v;
head->next = NULL;
}
else
{
BagNode * n = new BagNode; // new node
removePunct(v);
n->dataValue = v;
transform(v.begin(), v.end(), v.begin(), ::tolower);
n->dataCopy = v;
BagNode * current = head; //for traversal
//current = head;
n->dataCount = 0;
if(current->dataCopy > v)
{
n->next = head;
head = n;
}
else{ //mid and tail insert
while(current->next && current->next->dataCopy < v)
{
current = current->next;
}
n->next = current->next;
current->next = n;
}
}
BagNode * check = new BagNode;
for(check = head; check->next != NULL; check = check->next)
{
if(check->dataCopy == v)//isSame(check->dataValue, v))
{
check->dataCount++;
}
}
}
bool remove(string v) //bool
{
bool status;
if(head == NULL){
status = false;
}
else if(head->dataCopy > v)
{//(head->dataValue > v){
status = false;
}
else if(head->dataCopy == v)
{
BagNode * t = head;
head = head->next;
delete t;
status = true;
}
else//general case
{
BagNode * current = head;
while(current->next && current->next->dataCopy < v){
current = current->next;
}
if(current->next == NULL)
{
status = false;
}
else if(current->next->dataCopy == v) //found it
{
BagNode *t = current->next;
current->next = current->next->next;
delete t;
status = true;
}
else
{
status = false;
}
}
return status;
}
void traverse()
{
BagNode * current;
current = head;
while(current)
{
output << current->dataValue << " (" << current->dataCount << ")" << " ";
current = current->next;
}
cout << endl;
}
Output: 10Annette (1) 1805 (1) 7 (1) a (1) a (2) a (3) a (4) a (5) a (6) All (1) all (2) an (1) and (1) and (2) and (3) and (4) and (5) and (6) and (10) and (7)
if(!inputFile)
{
cout << "Could Not Open " << fileName << " File" << endl;
exit(EXIT_FAILURE);
}
else
{
while(inputFile >> text)
{
theBag.insert(text);
}
cout << "Processing File Complete" << endl;
cout << "Please Enter An Output File Name: ";
getline(cin,outputFilename);
output.open(outputFilename);
theBag.traverse();
theBag.remove(text);
inputFile.close();
output.close();
}
最佳答案
如果您在插入函数中查看此处,您实际上是在用该值接触每个节点。所以如果 v = "And"
每个单独的 "And"词都会增加它的数据计数。这会使您在每个节点上获得正确的单词计数。
for(check = head; check->next != NULL; check = check->next)
{
if(check->dataCopy == v)//isSame(check->dataValue, v))
{
check->dataCount++;
}
}
似乎您可以利用该行为使您的插入更简单。
关于c++ - 删除某些节点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40520003/