我已经问了一个与这个程序有关的问题，但现在经过大量研究和停止工作后，我再次陷入困境。
我正在尝试编写一个程序，该程序将接受用户输入并存储它然后打印出所有唯一的单词和它们每次发生的次数
例如
Please enter something: Hello#@Hello# hs,,,he,,whywhyto[then the user hits enter]
hello 2
hs 1
he 1
whywhyto 1
以上应该是输出，当然为什么不是一个字，但在这种情况下并不重要，因为我假设任何字母图案由任何非字母分隔（空格，0-9 ，＃$（@ etc.）被认为是一个单词。我需要使用2D数组，因为我无法使用链接列表，也不了解它们。
这就是我到目前为止所做的一切
#include <stdio.h>
#include <ctype.h>
int main()
{
char array[64];
int i=0, j, input;
printf("Please enter an input:");
input=fgetc(stdin);
while(input != '\n')
{
if(isalpha(input))
{
array[i]=input;
i++;
}
input=fgetc(stdin);
}
for(j=0;j<i;j++)
{
// printf("%c ",j,array[j]);
printf("%c",array[j]);
}
printf("\n");
}
我正在使用isalpha来获取字母，但所有这一切都是为了摆脱任何不是字母的东西，存储它然后打印回来，但我不知道如何获得它在第一次出现时存储一次单词，然后只为每个单词增加一个计数。我只能使用fgetc（），这对我来说至少很困难，我只有大约3-4个月的C经验，我知道我将不得不使用二维数组，已经阅读了它们但我还没有能够要理解我将如何实施它们，请帮我一点。
答案 0 :(得分：1)
不知道这是否是作业，所以我没有为你做任何事，我也清理了你的代码。但是，如果你不知道这个人可以输入多少单词，你需要一个动态的数据结构，如链表
#include <stdio.h>
#include <string.h>
#include <ctype.h>
typedef struct linkedlist linkedlist;
struct linkedlist{
char *word;
int count;
linkedlist *next;
};
int main()
{
//know your bounds, this will cause trouble if word is longer than 64 chars
char array[64];
int i=0, input;
linkedlist *head = NULL;
printf("Please enter an input:");
while((input=fgetc(stdin)) != '\n')
{
if(isalpha(input) && i!=63) //added this so that code does not brake (word is 64 chars)
{
array[i]=input;
}
else{
array[i]='\0';
char *word = malloc(strlen(array)+1);
strcpy(word, array);
add_word(word, &head);
i=0; //need to restart i to keep reading words
}
i++;
}
//print out final results
for(linkedlist *temp = head; temp != NULL; temp = temp->next){
printf("%s %d ", temp->word, temp->count);
}
}
//adds word to end of list if does not exist
//increments word count if it exists
void add_word(char *word, linkedlist **ll){
//implement this
}
//frees resources used by malloc (lookup how to free a linkedlist/destroy a linked list
//make sure to free both final and head in main
void destroy_list(linkedlist **ll){
//implement this
}
对于add_word，你需要的东西是（PSEUDO-CODE）：
list = *ll
if(list == NULL): //new list
*ll = malloc(sizeof(linkedlist))
ll->word = word
ll->count = 1
ll->next = NULL
return
while list->next != null:
if word = list->word:
free(word)
list->count++
return
list = list->next
if list->word = word: //last word in list
free(word)
list->count++
else: //word did not exist, add new word to end of list
temp = malloc(sizeof(linkedlist))
temp->word = word
temp->count = 1
list->next = temp
也许不是最有效的方法，但你可以改进它 希望我没有进一步混淆你，祝你好运
答案 1 :(得分：1)
以下是似乎有效的代码：
#include <assert.h>
#include <ctype.h>
#include <stdio.h>
#include <string.h>
enum { MAX_WORDS = 64, MAX_WORD_LEN = 20 };
int main(void)
{
char words[MAX_WORDS][MAX_WORD_LEN];
int count[MAX_WORDS] = { 0 };
int w = 0;
char word[MAX_WORD_LEN];
int c;
int l = 0;
while ((c = getchar()) != EOF)
{
if (isalpha(c))
{
if (l < MAX_WORD_LEN - 1)
word[l++] = c;
else
{
fprintf(stderr, "Word too long: %*s%c...\n", l, word, c);
break;
}
}
else if (l > 0)
{
word[l] = '\0';
printf("Found word <<%s>>\n", word);
assert(strlen(word) < MAX_WORD_LEN);
int found = 0;
for (int i = 0; i < w; i++)
{
if (strcmp(word, words[i]) == 0)
{
count[i]++;
found = 1;
break;
}
}
if (!found)
{
if (w >= MAX_WORDS)
{
fprintf(stderr, "Too many distinct words (%s)\n", word);
break;
}
strcpy(words[w], word);
count[w++] = 1;
}
l = 0;
}
}
for (int i = 0; i < w; i++)
printf("%3d: %s\n", count[i], words[i]);
return 0;
}
示例输出：
$ ./wordfreq <<< "I think, therefore I am, I think, or maybe I do not think after all, and therefore I am not."
Found word <<I>>
Found word <<think>>
Found word <<therefore>>
Found word <<I>>
Found word <<am>>
Found word <<I>>
Found word <<think>>
Found word <<or>>
Found word <<maybe>>
Found word <<I>>
Found word <<do>>
Found word <<not>>
Found word <<think>>
Found word <<after>>
Found word <<all>>
Found word <<and>>
Found word <<therefore>>
Found word <<I>>
Found word <<am>>
Found word <<not>>
5: I
3: think
2: therefore
2: am
1: or
1: maybe
1: do
2: not
1: after
1: all
1: and
$ ./wordfreq <<< "I think thereforeIamIthinkormaybeI do not think after all, and therefore I am not."
Found word <<I>>
Found word <<think>>
Word too long: thereforeIamIthinkor...
1: I
1: think
$ ./wordfreq <<< "a b c d e f g h i j k l m n o p q r s t u v w x y z
> A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
> aa ab ac ad ae af ag ah ai aj ak al am
> an ao ap aq ar as at au av aw ax ay az
> "
Found word <<a>>
Found word <<b>>
Found word <<c>>
Found word <<d>>
Found word <<e>>
Found word <<f>>
Found word <<g>>
Found word <<h>>
Found word <<i>>
Found word <<j>>
Found word <<k>>
Found word <<l>>
Found word <<m>>
Found word <<n>>
Found word <<o>>
Found word <<p>>
Found word <<q>>
Found word <<r>>
Found word <<s>>
Found word <<t>>
Found word <<u>>
Found word <<v>>
Found word <<w>>
Found word <<x>>
Found word <<y>>
Found word <<z>>
Found word <<A>>
Found word <<B>>
Found word <<C>>
Found word <<D>>
Found word <<E>>
Found word <<F>>
Found word <<G>>
Found word <<H>>
Found word <<I>>
Found word <<J>>
Found word <<K>>
Found word <<L>>
Found word <<M>>
Found word <<N>>
Found word <<O>>
Found word <<P>>
Found word <<Q>>
Found word <<R>>
Found word <<S>>
Found word <<T>>
Found word <<U>>
Found word <<V>>
Found word <<W>>
Found word <<X>>
Found word <<Y>>
Found word <<Z>>
Found word <<aa>>
Found word <<ab>>
Found word <<ac>>
Found word <<ad>>
Found word <<ae>>
Found word <<af>>
Found word <<ag>>
Found word <<ah>>
Found word <<ai>>
Found word <<aj>>
Found word <<ak>>
Found word <<al>>
Found word <<am>>
Too many distinct words (am)
1: a
1: b
1: c
1: d
1: e
1: f
1: g
1: h
1: i
1: j
1: k
1: l
1: m
1: n
1: o
1: p
1: q
1: r
1: s
1: t
1: u
1: v
1: w
1: x
1: y
1: z
1: A
1: B
1: C
1: D
1: E
1: F
1: G
1: H
1: I
1: J
1: K
1: L
1: M
1: N
1: O
1: P
1: Q
1: R
1: S
1: T
1: U
1: V
1: W
1: X
1: Y
1: Z
1: aa
1: ab
1: ac
1: ad
1: ae
1: af
1: ag
1: ah
1: ai
1: aj
1: ak
1: al
$
对“太长”字的测试＆＃39;和太多的单词＆＃39;帮助我保证代码是合理的。设计这样的测试是很好的做法。
答案 2 :(得分：0)
这个技巧是1）读取输入2）识别分隔符3）将单词与整个缓冲区进行比较，4）只打印一次。
这种方法具有内存效率，因为它只使用OP建议的64 char缓冲区。搜索复杂度为O（n * n）
#include <ctype.h>
#include <stdio.h>
#include <string.h>
// Helper function to find word occurrences.
void Print_count(const char *word, const char *array, int i) {
int count = 0;
const char *found;
for (int j = 0; j < i; j++) {
if (isalpha((unsigned char ) array[j])) {
if (strcmp(&array[j], word) == 0) {
found = &array[j];
count++;
}
// skip rest of word
do {
j++;
} while (isalpha((unsigned char ) array[j]));
}
}
if (found == word) {
printf("%s %d\n", word, count);
}
}
int main(void) {
char array[64];
int i = 0;
int j;
int input;
printf("Please enter an input:");
// get the input
while ((input = fgetc(stdin)) != '\n' && input != EOF) {
array[i] = input;
if (i + 1 >= sizeof array)
break;
i++;
}
array[i] = '\0';
// change all delimiters to \0
for (j = 0; j < i; j++) {
if (!isalpha((unsigned char ) array[j])) {
array[j] = '\0';
}
}
for (j = 0; j < i; j++) {
// Use the beginning of each word ...
if (isalpha((unsigned char ) array[j])) {
Print_count(&array[j], array, i);
// skip test of word
do {
j++;
} while (isalpha((unsigned char ) array[j]));
}
}
return 0;
}
输入Hello#@Hello# hs,,,he,,whywhyto
输出：
Hello 2
hs 1
he 1
whywhyto 1