我遇到以下问题：我需要输入单词直到EOF，然后再输入由完全相同的字符（不一定是完全相同的字符）组成的单词。
例如：
输入：
"abc" "acb" "abcabc" "cab" "de" "gh" "ab" "ed" "hg" "abcde"
输出：
"abc" "acb" "abcabc" "cab"
"de" "ed"
"gh" "hg"
我将放下到目前为止所做的事情：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char **groups, word[30], **store, n = 0, k = 0;
store = malloc(1*sizeof(char *));
if (store == NULL) exit(1);
for (;;) {
printf("Enter word: ");
if (scanf("%s", word) == EOF) break;
store[n] = malloc((strlen(word)+1)*sizeof(char));
if (store[n] == NULL) exit(1);
strcpy(store[n], word);
n++;
store = realloc(store, (n+1)*sizeof(char *));
if (store == NULL) exit(1);
}
for (int i=0; i<n; i++) {
printf("%s ", store[i]);
}
return 0;
}
问题是我真的不知道如何检查字符。你能帮我吗？
更新
我尝试按照@jarmod的建议进行操作：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char **groups, word[30], **store, n = 0, k = 0, *aux;
store = malloc(1*sizeof(char *));
if (store == NULL) exit(1);
for (;;) {
printf("Enter word: ");
if (scanf("%s", word) == EOF) break;
store[n] = malloc((strlen(word)+1)*sizeof(char));
if (store[n] == NULL) exit(1);
strcpy(store[n], word);
n++;
store = realloc(store, (n+1)*sizeof(char *));
if (store == NULL) exit(1);
}
for (int i=0; i<n; i++) {
printf("%s ", store[i]);
}
printf("\n");
for (int i=0; i<n; i++) {
for (int j=0; j<strlen(store[i])-1; j++) {
for (int l=(j+1); l<strlen(store[i]); l++) {
if (store[i][j] > store[i][l]) {
aux = store[i][j];
store[i][j] = store[i][l];
store[i][l] = aux;
}
}
}
}
for (int i=0; i<n; i++) {
printf("%s ", store[i]);
}
printf("\n");
for (int i=0; i<n; i++) {
for (int j=0; j<strlen(store[i])-1; j++) {
if (store[i][j] == store[i][j+1]) {
for (int l=j; l<strlen(store[i])-1; l++) {
store[i][l] = store[i][l+1];
}
j--;
store[i] = realloc(store[i], (strlen(store[i])-1)*sizeof(char));
if (store[i] == NULL) exit(1);
}
}
}
for (int i=0; i<n; i++) {
printf("%s ", store[i]);
}
printf("\n");
return 0;
}
答案 0 :(得分：0)
我的方法是为每个单词使用26位掩码，以便您可以按掩码值对它们进行分组。
我已经实现了，但是实际的分组和显示部分尚未优化。我确实解决了您的难题难题
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
int matched = 0;
int n_words = 0;
char** words = NULL;
int* masks = NULL;
int max_mask = 0;
while (1) { // Read until EOF breaks
char word[30];
if (scanf("%s\n", word) == EOF)
break;
words = realloc(words, sizeof(char *) * (n_words + 1));
words[n_words] = strndup(word, 30);
n_words++;
}
// Computes a bit mask for each word
masks = malloc(sizeof(int) * n_words);
for (int i = 0; i < n_words; ++i) {
masks[i] = 0;
for (int j = 0; words[i][j] != '\0'; ++j) {
int bit_index = words[i][j] - 'a';
masks[i] |= (1 << bit_index);
}
if (masks[i] > max_mask)
max_mask = masks[i];
}
// Iterate over each possible masks and print all words with this mask
for(int i = 0; i <= max_mask; i++) {
for (int w = 0; w < n_words; w++) {
if (masks[w] == i) {
printf("%s ", words[w]);
matched = 1;
}
}
if (matched) {
printf("\n");
matched = 0;
}
}
return 0;
}
答案 1 :(得分：0)
有一个聪明的方法来确定一个单词属于哪个组。
我们必须创建一个方案，并用数字唯一标识每个组。我们如何定义和区分组：通过其包含的字母。重复字母不计算在内（这是一组）。因此，我们需要一个公式来编码一组字母。如果我们为每个字母分配一个数字（从0开始，我们将通过2的幂的和来获得id（实际上可以选择任何基数，但是2是计算的自然选择）。
例如：
"abdeae"字母集合为{'a', 'b', 'd', 'e'}。它们对应的编号为：{0, 1, 3, 4}，id为2^0 + 2^1 + 2^3 + 2^4。
因为有26个字母，所以我们可以使用32位整数来编码ID。 2^i对应于位i，因此算法看起来像这样：
uint32_t letter_mask(char ch)
{
assert(ch >= 'a' && ch <= 'z');
return (uint32_t) 1u << (ch - 'a');
}
uint32_t word_group_id(const char * str)
{
size_t len = strlen(str);
uint32_t id = 0;
for (size_t i = 0; i < len; ++i)
{
id |= letter_mask(str[i]);
}
return id;
}
现在，我们有了一种简单的方法来确定一个单词的组，您需要创建一个简化版本的地图以将单词放在其中。
这是我简单的快速实现。免责声明：未经测试。您还必须通过添加对malloc和realloc的检查来改进它。
typedef struct Word_map_bucket
{
uint32_t id;
char** words;
size_t words_size;
} Word_map_bucket;
void init_word_map_bucket(Word_map_bucket* bucket, uint32_t id)
{
bucket->id = id;
bucket->words = NULL;
bucket->words_size = 0;
}
typedef struct Word_map
{
Word_map_bucket* buckets;
size_t buckets_size;
} Word_map;
void init_word_map(Word_map* map)
{
map->buckets = NULL;
map->buckets_size = 0;
}
Word_map_bucket* find_bucket(Word_map map, uint32_t id)
{
for (size_t i = 0; i < map.buckets_size; ++i)
{
if (map.buckets[i].id == id)
return &map.buckets[i];
}
return NULL;
}
Word_map_bucket* add_new_bucket(Word_map* map, uint32_t id)
{
map->buckets = realloc(map->buckets, map->buckets_size + 1);
map->buckets_size += 1;
Word_map_bucket* bucket = &map->buckets[map->buckets_size + 1];
init_word_map_bucket(bucket, id);
return bucket;
}
void add_word(Word_map* map, const char* word)
{
// get to bucket
uint32_t id = word_group_id(word);
Word_map_bucket* bucket = find_bucket(*map, id);
if (bucket == NULL)
bucket = add_new_bucket(map, id);
// increase bucket->words
bucket->words = realloc(bucket->words, bucket->words_size + 1);
bucket->words_size += 1;
// push word into bucket
bucket->words[bucket->words_size - 1] = malloc(strlen(word));
strcpy(bucket->words[bucket->words_size - 1], word);
}
答案 2 :(得分：-1)
int count[256]获取每个字符是否在每个单词的单词中，例如，对于"abcabc"，使count['a'] = 1 count['b'] = 1 count['c'] = 1 qsort具有特殊的比较用途count count相同，则输出，如果不输出下一行，则输出一行。以下code可以工作：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Node {
unsigned char* st;
int count[256];
};
int compare(const void* x, const void* y) {
const struct Node* node_x = (const struct Node*)x;
const struct Node* node_y = (const struct Node*)y;
for (int i = 0; i != 256; ++i) {
if (node_x->count[i] > node_y->count[i])
return -1;
if (node_x->count[i] < node_y->count[i])
return 1;
}
return 0;
}
int main() {
unsigned char f[][10] = {"abc", "acb", "abcabc", "cab", "de", "gh", "ab", "ed", "hg"};
int n = sizeof(f) / sizeof(f[0]);
struct Node node[100];
for (int i = 0; i != n; ++i) {
node[i].st = f[i];
for (int j = 0; j != sizeof(node[i].count) / sizeof(node[i].count[0]); ++j)
node[i].count[j] = 0;
for (int j = 0; f[i][j] != '\0'; ++j) {
unsigned char ch = f[i][j];
node[i].count[ch] = 1;
}
}
qsort(node, n, sizeof(struct Node), compare);
int t = 0;
for (int i = 0; i < n; ++i) {
if (i == 0) {
++t;
continue;
}
int k = i - 1;
if (memcmp(node[i].count, node[k].count, sizeof(node[i].count)) == 0) {
if (t == 1)
printf("%s ", node[k].st);
printf("%s ", node[i].st);
++t;
} else {
if (t != 0)
printf("\n");
t = 0;
}
}
printf("\n");
return 0;
}