我使用了分隔符
strtok(string, ",.;:\"!? -_\n\t*()@#=+");
并将其保存在
中char *temp[100];
现在我如何检查单词的重复？并显示如下
3 won't
2 talk
1 can't
1 walk
它应显示从最高重复到最低。如果重复相同则显示字母顺序。
抱歉我的英语不好。
答案 0 :(得分：3)
使用std :: string来保存strtok()的结果。然后创建一个std::map<string, int>来保存字符串（密钥）发生的次数。
您可以使用以下内容填充地图：
std::map<string, int> myMap;
myMap[tokenizedWord]++; //Increase count of word.
然后，您可以循环浏览地图内容，并在整数值大于2的任何地方打印出来。
for (std::map<string, int>::iterator iter = myMap.begin(); iter != myMap.end(); ++iter)
{
if (iter->second > 1)
std::cout << "Duplicated word: " << iter->first << " count = " << iter->second;
}
我会让你弄清楚如何按顺序遍历它。您可以将值放在矢量或其他内容中，然后在打印之前使用std::sort或其他任何您喜欢的内容。不幸的是，地图是关联容器，你不能对它们进行排序，因为它会破坏它们的内部排序。
std :: map的背景信息
映射是一个关联数组，意味着每个键都映射到特定值，并且键是唯一的。实际上，您可以创建一个多重映射，其中键不是唯一的，因此这很重要。
基本上，由于密钥是唯一的，因此只需将密钥用作数组索引即可访问或创建元素。
例如：
//Create a map and insert a couple things into it - prices of meat?
std::map<string, float> myMap;
myMap["Chicken"] = 4.99;
myMap["Turkey"] = 6.99;
//Retrieve the price of something using the key.
std::cout << "Chicken costs " << myMap["Chicken"] << std::end;
您也可以在地图上执行标准插入和定位操作，但关联数组语法更简单，为什么还要烦恼呢？ ：）
PS：要完全回答您的评论，以防万一，myMap [tokenizedWord] ++末尾的++只是说增加为该键存储的整数值的值你可以做myMap [tokenizedWord] = myMap [tokenizedWord] + 1或者你也可以做myMap [tokenizedWord] + = 1。
答案 1 :(得分：1)
完整实现您的问题（如果您想要一个示例代码进行排序，请告诉我们）：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#define ARRAY_ELEMS_COUNT(A) sizeof(A)/sizeof(*A)
typedef struct _word_t
{
char *word;
int occurr_count;
struct _word_t *next;
} word_t;
typedef struct _word_list_t
{
struct _word_t *head;
struct _word_t *tail;
int elems_count;
} word_list_t;
/* Creation of the words list */
word_list_t *make_list(void)
{
word_list_t *w_list = (word_list_t *)malloc(sizeof (struct _word_list_t));
if (w_list == NULL)
{
fprintf(stderr, "malloc faild --> %s\n", strerror(errno));
return NULL;
}
w_list->head = w_list->tail = NULL;
w_list->elems_count = 0;
return w_list;
}
int list_word_lookup(word_list_t *w_list, char *word)
{
word_t *temp_word = w_list->head;
while(temp_word)
{
if (strcmp(temp_word->word, word) == 0)
{
/* We got it before, increment the count */
temp_word->occurr_count++;
return 1;
}
else
{
temp_word = temp_word->next;
}
}
return 0;
}
/* Adding new words to the list of words if they are not present, otherwise increment their occurrence count */
/* TODO : Sort the list using Merge sort for performance */
int adding_to_list(word_list_t *w_list, char *word)
{
int return_status = 0;
char *tmp_word = (char *)malloc(sizeof(char)*(strlen(word) + 1));
word_t *new_word = (word_t *)malloc(sizeof(struct _word_t));
/* Empty list */
if (w_list->head == NULL)
{
strcpy(tmp_word, word);
new_word->word = tmp_word;
new_word->occurr_count = 1;
w_list->head = w_list->tail = new_word;
w_list->head->next = NULL;
w_list->elems_count++;
}
else
{
/* The list is not empty */
/* Checking if the word exist in the list */
return_status = list_word_lookup(w_list, word);
if (return_status == 1)
{
fprintf(stdout, "WE got this word before --> increment count\n");
}
else
{
strcpy(tmp_word, word);
new_word->word = tmp_word;
new_word->occurr_count = 1;
w_list->tail->next = new_word;
w_list->tail = new_word;
w_list->tail->next = NULL;
}
}
return 0;
}
void words_list_dump(word_list_t *w_list)
{
word_t *temp;
for (temp = w_list->head; temp; temp = temp->next) {
fprintf(stdout, "Word : %s -- Count = %d\n", temp->word, temp->occurr_count);
}
}
/* Destroying all words */
void free_words(word_list_t *w_list)
{
word_t *temp;
for (temp = w_list->head; temp; temp = temp->next) {
/* Freeing the word string */
free(temp->word);
/* Freeing the word */
free(temp);
}
w_list->head = NULL;
w_list->tail = NULL;
}
/* Destroying the words list */
void free_words_list(word_list_t *w_list)
{
if (!w_list)
{
return;
}
free_words(w_list);
free(w_list);
}
/* TODO : create a function that converts your input text to a char ** array, so you can pass it to adding_to_list */
/* For testing */
int main(int argc, char **argv)
{
const char *string[] = {"Hello", "World", "Stackoverflow", "C", "Hello", "C", "WORDS", "words", "List", "list", "Hello", "World", "Count"};
word_list_t *my_list = make_list();
int i;
for (i = 0; i < ARRAY_ELEMS_COUNT(string); i++)
adding_to_list(my_list, string[i]);
words_list_dump(my_list);
free_words_list(my_list);
return 0;
}
答案 2 :(得分：0)
以下是使用strtok但没有std::map的答案。在一个字符串传递中，每个单词都与前一个单词进行检查，并重复计算。
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
#include <string>
using std::string;
#include <vector>
using std::vector;
#include <cstring>
using std::tolower;
int main()
{
char *strin;
string inputstr;
vector<string> svec;
vector<int> cvec;
char *pch;
int unique_word_count=0;
while(getline(cin,inputstr))
{
//token-ize the string
//First string
strin = &inputstr[0];
pch = std::strtok(strin," ,-");
bool unique_word_found = true;
//subsequent words
while (pch != NULL)
{
string word(pch);
for(string::size_type i=0; i < word.size(); i++)
word[i]=tolower(word[i]);
//first word
//just add to svec and no comparisons
if(unique_word_count==0)
{
svec.push_back(word);
cvec.push_back(1);
cvec[unique_word_count++]=1; //init count of first word
//next word
pch = std::strtok(NULL, " ,-");
unique_word_found = true; //reset flag
continue;
}
//start comparing with other words currently in string vector
//do not do this if only 1 word present
vector<string>::iterator iter=svec.begin();
while(iter < svec.end())
{
if(word == *iter)
{
//match found
cvec[iter-svec.begin()]++; //increment count of that word
unique_word_found = false;
}
iter++;
}
if(unique_word_found)
{
//add to unique word list and increment count
svec.push_back(word);
cvec.push_back(1);
cvec[unique_word_count++]=1;
}
//next word
pch = std::strtok(NULL, " ,-");
unique_word_found = true; //reset flag
}
}
cout << "Word" << " ---> " << "Occurences" << endl;
for(vector<string>::size_type i=0; i < svec.size(); i++)
{
cout << svec[i] << " ---> " << cvec[i] << endl;
}
return 0;
}
答案 3 :(得分：0)
总体策略如下：