如何忽略空格和标点符号?
How to ignore whitespace and punctuation?
我写了一个回文检查器函数,它在大多数情况下都有效,但如果空格或标点符号不在字符串的中间,它就说它不是回文。
即
第一次测试:
Enter string to test for palindrome:
hannah
string is a palindrome.
第二次测试:
Enter string to test for palindrome:
han nah
string is a palindrome.
第三次测试:
Enter string to test for palindrome:
hann.ah
string is not a palindrome.
第四次测试:
Enter string to test for palindrome:
han.nah
string is a palindrome.
我想知道是否有一种方法可以同时忽略空格和标点符号,以便将 h.annah
或 hann ah
视为回文?
这是我的代码:
void isPalindrome (string s){
if(equal(s.begin(), s.begin() + s.size()/2, s.rbegin()) )
cout << "string is a palindrome. " << endl;
else
cout << "string is not a palindrome. " << endl;
}
int main(){
string test1;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test1);
isPalindrome(test1);
string test2;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test2);
isPalindrome(test2);
string test3;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test3);
isPalindrome(test3);
return 0;
}
在回文检查之前对字符串应用过滤器。
这是一种方法。
#include <string>
#include <iostream>
#include <algorithm>
void isPalindrome (std::string s){
if(equal(s.begin(), s.begin() + s.size()/2, s.rbegin()) )
std::cout << "string is a palindrome. " << std::endl;
else
std::cout << "string is not a palindrome. " << std::endl;
}
std::string remove_rubbish(std::string s)
{
auto is_rubbish = [](char c)
{
return std::ispunct(c) || std::isspace(c);
};
s.erase(std::remove_if(s.begin(),
s.end(),
is_rubbish),
s.end());
return s;
}
int main(){
auto s= std::string("ha-n.n?a h");
isPalindrome(remove_rubbish(s));
return 0;
}
没问题!只需定义 C++ 标准中尚未定义的算法 equal_if
。:)
这是一个演示程序
#include <iostream>
#include <string>
#include <cctype>
template <typename InputIterator1, typename InputIterator2, typename UnaryPredicate>
bool equal_if(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2,
UnaryPredicate unary_predicate)
{
do
{
while (first1 != last1 && !unary_predicate(*first1)) ++first1;
while (first2 != last2 && !unary_predicate(*first2)) ++first2;
} while (first1 != last1 && first2 != last2 && *first1++ == *first2++);
return first1 == last1 && first2 == last2;
}
int main()
{
std::string s1("h.annah");
std::string s2("hann ah");
if (equal_if(s1.begin(), s1.end(), s1.rbegin(), s1.rend(), ::isalpha))
{
std::cout << "The string \"" << s1 << "\" is a palindrome" << std::endl;
}
if (equal_if(s2.begin(), s2.end(), s2.rbegin(), s2.rend(), ::isalpha))
{
std::cout << "The string \"" << s2 << "\" is a palindrome" << std::endl;
}
return 0;
}
它的输出是
The string "h.annah" is a palindrome
The string "hann ah" is a palindrome
我写了一个回文检查器函数,它在大多数情况下都有效,但如果空格或标点符号不在字符串的中间,它就说它不是回文。
即
第一次测试:
Enter string to test for palindrome:
hannah
string is a palindrome.
第二次测试:
Enter string to test for palindrome:
han nah
string is a palindrome.
第三次测试:
Enter string to test for palindrome:
hann.ah
string is not a palindrome.
第四次测试:
Enter string to test for palindrome:
han.nah
string is a palindrome.
我想知道是否有一种方法可以同时忽略空格和标点符号,以便将 h.annah
或 hann ah
视为回文?
这是我的代码:
void isPalindrome (string s){
if(equal(s.begin(), s.begin() + s.size()/2, s.rbegin()) )
cout << "string is a palindrome. " << endl;
else
cout << "string is not a palindrome. " << endl;
}
int main(){
string test1;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test1);
isPalindrome(test1);
string test2;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test2);
isPalindrome(test2);
string test3;
cout << "Enter string to test for palindrome: " << endl;
getline(cin, test3);
isPalindrome(test3);
return 0;
}
在回文检查之前对字符串应用过滤器。
这是一种方法。
#include <string>
#include <iostream>
#include <algorithm>
void isPalindrome (std::string s){
if(equal(s.begin(), s.begin() + s.size()/2, s.rbegin()) )
std::cout << "string is a palindrome. " << std::endl;
else
std::cout << "string is not a palindrome. " << std::endl;
}
std::string remove_rubbish(std::string s)
{
auto is_rubbish = [](char c)
{
return std::ispunct(c) || std::isspace(c);
};
s.erase(std::remove_if(s.begin(),
s.end(),
is_rubbish),
s.end());
return s;
}
int main(){
auto s= std::string("ha-n.n?a h");
isPalindrome(remove_rubbish(s));
return 0;
}
没问题!只需定义 C++ 标准中尚未定义的算法 equal_if
。:)
这是一个演示程序
#include <iostream>
#include <string>
#include <cctype>
template <typename InputIterator1, typename InputIterator2, typename UnaryPredicate>
bool equal_if(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2,
UnaryPredicate unary_predicate)
{
do
{
while (first1 != last1 && !unary_predicate(*first1)) ++first1;
while (first2 != last2 && !unary_predicate(*first2)) ++first2;
} while (first1 != last1 && first2 != last2 && *first1++ == *first2++);
return first1 == last1 && first2 == last2;
}
int main()
{
std::string s1("h.annah");
std::string s2("hann ah");
if (equal_if(s1.begin(), s1.end(), s1.rbegin(), s1.rend(), ::isalpha))
{
std::cout << "The string \"" << s1 << "\" is a palindrome" << std::endl;
}
if (equal_if(s2.begin(), s2.end(), s2.rbegin(), s2.rend(), ::isalpha))
{
std::cout << "The string \"" << s2 << "\" is a palindrome" << std::endl;
}
return 0;
}
它的输出是
The string "h.annah" is a palindrome
The string "hann ah" is a palindrome