Leetcode--valid palindrome (palindrome judgment)

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:
Are you consider that the string might is empty? This was a good question to ask during a interview.

For the purpose of this problem, we define empty string as valid palindrome.

Remove extraneous characters from S, convert uppercase letters to lowercase, then simple palindrome judgments.

Written in the C language version, can achieve the effect, but will prompt timeout

BOOLIspalindrome (Char*s) {intj =0;  for(intI=0; I<strlen (s); i++){        if(s[i]>='0'&& s[i]<='9'|| s[i]>='a'&& s[i]<='Z'|| s[i]>='A'&& s[i]<='Z'){            if(s[i]>='A'&& s[i]<='Z') {s[j+ +] = s[i]-'A'+'a'; }Else{s[j++] =S[i]; }}} S[j]=' /'; if(J <=1)return true;  for(intI=0; I<strlen (s); i++){        if(S[i]! = S[strlen (s)-i-1]){            return false; }    }    return true;}

Online to find the C + + version, verify that you can use

classSolution { Public:    BOOLIspalindrome (strings) {stringt ="";  for(inti =0; I < s.length (); i++) {              if(S[i] >='a'&& S[i] <='Z'|| S[i] >='0'&& S[i] <='9'|| S[i] >='A'&& S[i] <='Z') {                  if(S[i] >='A'&& S[i] <='Z') {T+ = (S[i]-'A'+'a'); }                  Else{T+=S[i]; }              }          }          if(T = ="") {              return true; }           for(inti =0; I < t.length ()/2; i++) {              if(T[i]! = t[t.length ()-I-1]) {                  return false; }          }          return true; }};

Leetcode--valid palindrome (palindrome judgment)