Leetcode[string]: Valid palindrome

given A string, determine if it is a palindrome, considering only alphanumeric characters and Igno Ring CASES. 
for example, 
" a man, a plan, a Canal:panama "  is a palindrome. 
" race a car "  is not a palindrome.

This problem is relatively simple, the general idea solution is as follows:

    BOOLIspalindrome (strings) { for(inti =0, j = s.size ()-1; I < J; ++i,--j) { while(I < s.size ()) {if(S[i] >=' 0 '&& S[i] <=' 9 ') Break;Else if(S[i] >=' A '&& S[i] <=' Z ') Break;Else if(S[i] >=' A '&& S[i] <=' Z ') {S[i] + = (' A '-' A '); Break;            } ++i; } while(J >=0)            {if(S[j] >=' 0 '&& S[j] <=' 9 ') Break;Else if(S[j] >=' A '&& S[j] <=' Z ') Break;Else if(S[j] >=' A '&& S[j] <=' Z ') {S[j] + = (' A '-' A '); Break;            }--j; }if(S[i]! = S[j])return false; }return true; }

Another solution is to use the Isalnum () and ToLower () functions in C + +. The descriptions of the two functions are as follows:

Isalnum ():

ToLower ():

The C + + solution using these two functions is as follows:

    BOOLIspalindrome (strings) { for(inti =0, j = s.size ()-1; I < J; ++i,--j) { while(I < S.size () &&!isalnum(S[i])) ++i; while(J >=0&&!isalnum(S[j])) --j;if(ToLower(S[i])! =ToLower(S[j]))return false; }return true; }

The time complexity of both methods is O (N), the spatial complexity is O (1), the time performance is similar, as shown in:

In addition, the problem can be solved using regular expressions.

Leetcode[string]: Valid palindrome