我有两个字符串,我想知道它们是否相等。但是字符串字符顺序已经随机化。某些字符也可能已使用通配符运算符(*)进行换出。我用它来进行Anagram检测。
在这种情况下,我试图获得anagram程序,我不得不说ab **是abba的字谜。现在它可以判断它是否是一个字谜,如果它实际上是像阿巴和bbaa这样的字谜。现在我想弄清楚如何实现通配符*,我不知道从哪里开始,请帮忙!
到目前为止:
#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#define SIZE 5
bool areAnagram(char *str1, char *str2)
{
int count[SIZE] = {0};
int i = 0;
for (i = 0; str1[i] && str2[i]; i++)
{
count[str1[i]]++;
count[str2[i]]--;
}
if (str1[i] || str2[i])
{
return false;
}
for (i = 0; i < SIZE; i++)
{
if (count[i])
{
return false;
}
}
return true;
}
int main()
{
char str1[SIZE], str2[SIZE];
FILE *finput;
finput = fopen("input.txt", "r");
fscanf(finput, "%s %s", str1, str2);
printf("%s\n", str1);
printf("%s\n", str2);
if(areAnagram(str1, str2))
{
printf("THEY ARE ANAGRAMS\n");
}
else
{
printf("THEY AREN'T ANAGRAMS\n");
}
}
答案 0 :(得分:1)
类似于选择排序算法的工作方式,但不是检查整数值来排列我只是使用字符串比较来选择/删除并为(*)添加一个例外。迭代整个列表和中提琴!
#include <stdio.h>
#include <string.h>
int main() {
char inputString[] = "AB**";//input string (with asterisks)
char comparisonString[] = "AYYB";//comparison string (without asterisks)
int inputString_length = strlen(inputString);
int comparisonString_length = strlen(comparisonString);
int anagram = 1;//boolean
if (inputString_length != comparisonString_length) {
anagram = 0;
} else {
int i = 0;
while ((i < inputString_length) && (anagram == 1)) {
char *letterToCheck = inputString[0];
memmove(&inputString[0], &inputString[0 + 1], strlen(inputString) - 0);//remove first character
int j = 0;
int comparisonString_length_new = strlen(comparisonString);
int matchFound = 0;//boolean
while ((j < comparisonString_length_new) && (matchFound == 0)) {
char *letterToCompare = comparisonString[j];
if (letterToCheck == '*') {
matchFound = 1;
}
if (letterToCheck == letterToCompare) {
matchFound = 1;
memmove(&comparisonString[j], &comparisonString[j + 1], strlen(comparisonString) - j);//remove matched character
}
j++;
}
if (matchFound == 0) {
anagram = 0;
}
i++;
}
}
if (anagram == 0) {
printf("Are NOT Anagrams");
} else {
printf("Are Anagrams");
}
}
输入和比较字符串可能包含*的另一种解决方案(注意:下面的解决方案是由非现场用户添加而不是OP,它也未经OP测试或验证)
#include <stdio.h>
#define CHAR_LEN 4
#define SPECIAL_CHARACTER '*'
int count_special_char(char *string) {
int i = 0, count = 0;
for(i = 0; i < CHAR_LEN; i++) {
if(string[i] == SPECIAL_CHARACTER)
count++;
}
return count;
}
int is_anagram(char *string_a, char *string_b) {
int i, y;
int found_count = 0;
int a_special = count_special_char(string_a);
int b_special = count_special_char(string_b);
for(i = 0; i < CHAR_LEN; i++) {
if(string_a[i] == SPECIAL_CHARACTER) //compare only non-asterisk char
continue;
for(y = 0; y < CHAR_LEN; y++) {
if(string_a[i] == string_b[y]) {
string_b[y] = '\0' //treat this char as found
found_count++;
break;
}
}
}
if((found_count + a_special + b_special) >= CHAR_LEN)
return 1;
else
return 0;
}
int main() {
char a[CHAR_LEN] = "**CD";
char b[CHAR_LEN] = "AB**";
if(is_anagram(a, b))
printf("yes\n");
else
printf("no\n");
return 0;
}
A ***和* XYZ被假定为字谜,因为第一个字符串有3 *,它可以代表第二个字符串XYZ。第二个字符串有1 *,它可以代表第一个字符串A.如果有任何错误,请指出并提供帮助。谢谢!
答案 1 :(得分:-1)
在Linux和Posix系统上,您可以使用globbing相关函数。请参阅glob(7)并查看glob(3),fnmatch(3),wordexp(3)等
同时查看regcomp(3)
如果您需要自己撰写,请参阅finite state machines,context free grammars,regular grammars,regular expressions,parsing。