题目:
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,"A man, a plan, a canal: Panama"
is a palindrome."race a car"
is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
代码:
class Solution {
public:
bool isPalindrome(string s) {
std::transform(s.begin(), s.end(), s.begin(),::tolower);
std::string::iterator begin = s.begin();
std::string::iterator end = s.end();
while ( begin < end )
{
if ( !::isalnum(*begin) ){
++begin;
}
else if ( !::isalnum(*end) ){
--end;
}
else if ( *begin != *end ){
return false;
}
else{
++begin;
--end;
}
}
return true;
}
};
Tips:
1. isalnum transform函数省去了不少篇幅
2. 双指针技巧,从两头往中间逼近,不用判断奇数偶数,代码很简洁。
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第二次过回文判断的题,大体思路还在,iswalnum和transform能想起来有这么个东西,具体用法记不住了。代码改了一次以后AC了。
class Solution {
public:
bool isPalindrome(string s) {
if (s.size()==) return true;
std::transform(s.begin(), s.end(), s.begin(),::tolower);
int p1 = ;
int p2 = s.size()-;
while ( p1<p2 )
{
if ( !::iswalnum(s[p1]) ) { p1++; continue; }
if ( !::iswalnum(s[p2]) ) { p2--; continue; }
if ( s[p1++]!=s[p2--] ) return false;
}
if (p1>p2) return true;
return s[p1]==s[p2];
}
};