我必须修改我的程序以接受来自
名为anagrams.txt的文件。该文件每行应有两个字符串,用字符分隔我的程序应该是
每对字符串,如果每对字符串是一个anagram,则返回报告例如,考虑anagrams.txt的以下内容:
你好#elloh
曼南
阿童木
您的程序应打印出以下内容:
你好#elloh-Anagrams!
男#南-字谜!
阿童木-不是字谜!
我应该编译成g++
下面是从文本中读取的代码:
int main()
{
char input[30];
if(access( "anagrams.txt", F_OK ) != -1) {
FILE *ptr_file;
char buf[1000];
ptr_file =fopen("anagrams.txt","r"); if (!ptr_file)
return 1;
while (fgets(buf,1000, ptr_file)!=NULL)
printf("%s",buf);
fclose(ptr_file);
printf("\n");
}
else{ //if file does not exist
printf("\nFile not found!\n");
}
return 0;
}
查找文本是否为anagrams的代码:
#include <stdio.h>
int find_anagram(char [], char []);
int main()
{
char array1[100], array2[100];
int flag;
printf("Enter the string\n");
gets(array1);
printf("Enter another string\n");
gets(array2);
flag = find_anagram(array1, array2);
if (flag == 1)
printf(" %s and %s are anagrams.\n", array1, array2);
else
printf("%s and %s are not anagrams.\n", array1, array2);
return 0;
}
int find_anagram(char array1[], char array2[])
{
int num1[26] = {0}, num2[26] = {0}, i = 0;
while (array1[i] != '\0')
{
num1[array1[i] - 'a']++;
i++;
}
i = 0;
while (array2[i] != '\0')
{
num2[array2[i] -'a']++;
i++;
}
for (i = 0; i < 26; i++)
{
if (num1[i] != num2[i])
return 0;
}
return 1;
}
最佳答案
你可以试试这样的方法:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define MAXLINE 1000
#define MAXLETTER 256
int is_anagram(char *word1, char *word2);
void check_lines(FILE *filename);
int cmpfunc(const void *a, const void *b);
void convert_to_lowercase(char *word);
int
main(int argc, char const *argv[]) {
FILE *filename;
if ((filename = fopen("anagram.txt", "r")) == NULL) {
fprintf(stderr, "Error opening file\n");
exit(EXIT_FAILURE);
}
check_lines(filename);
fclose(filename);
return 0;
}
void
check_lines(FILE *filename) {
char line[MAXLINE];
char *word1, *word2, *copy1, *copy2;
while (fgets(line, MAXLINE, filename) != NULL) {
word1 = strtok(line, "#");
word2 = strtok(NULL, "\n");
copy1 = strdup(word1);
copy2 = strdup(word2);
convert_to_lowercase(copy1);
convert_to_lowercase(copy2);
if (is_anagram(copy1, copy2)) {
printf("%s#%s - Anagrams!\n", word1, word2);
} else {
printf("%s#%s - Not Anagrams!\n", word1, word2);
}
}
}
void
convert_to_lowercase(char *word) {
int i;
for (i = 0; word[i] != '\0'; i++) {
word[i] = tolower(word[i]);
}
}
int
is_anagram(char *word1, char *word2) {
qsort(word1, strlen(word1), sizeof(*word1), cmpfunc);
qsort(word2, strlen(word2), sizeof(*word2), cmpfunc);
if (strcmp(word1, word2) == 0) {
return 1;
}
return 0;
}
int
cmpfunc(const void *a, const void *b) {
if ((*(char*)a) < (*(char*)b)) {
return -1;
}
if ((*(char*)a) > (*(char*)b)) {
return +1;
}
return 0;
}
关于c - 从文本文件中读取,并使用每一行进行比较,看看它们是否是字谜,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40235955/