LeetCode——Valid Palindrome (回文判断)

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.

去掉S中的无关字符，将大写字母转换为小写，之后就是简单的回文串判断。

自己写的C语言版的，可以实现效果，但是会提示超时

bool isPalindrome(char *s) {
    int j = 0;
    for(int i=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] = ‘\0‘;
    if(j <= 1) return true;
    for(int i=0; i<strlen(s); i++){
        if(s[i] != s[strlen(s)-i-1]){
            return false;
        }
    }
    return true;
}

网上找的C++版的，验证可以使用

class Solution {
public:
    bool isPalindrome(string s) {
        string t = "";
        for (int i = 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 (int i = 0; i < t.length() / 2; i++) {
            if (t[i] != t[t.length() - i - 1]) {
                return false;
            }
        }
        return true;
    }
};