[leetcode]Valid Palindrome

相对简单，就是从两边往中间推进，忽略其他字符。

public class Solution {
    public boolean isPalindrome(String s) {
        if (s.length() == 0) return true;
        int i = 0;
        int j = s.length() - 1;
        while (true)
        {
            while (i < s.length() && !isAlphaNum(s.charAt(i)))
            {
                i++;
            }
            while (j >= 0 && !isAlphaNum(s.charAt(j)))
            {
                j--;
            }
            if (i > j || i == s.length() || j < 0) break;
            int c1 = s.charAt(i);
            int c2 = s.charAt(j);
            if (c1 == c2 || c1-c2 == 'a'-'A' || c1-c2 == 'A' - 'a') {
                i++;
                j--;
            }
            else return false;
        }
        return true;
    }
    
    private boolean isAlphaNum(char c)
    {
        if ((c >= 'a' && c <= 'z') ||
            (c >= 'A' && c <= 'Z') ||
            (c >= '0' && c <= '9')) return true;
        else return false;
    }
}

这个参考答案写的很简洁，当然它修改了原来的字符串，不过这不是关键，也可以不改，关键是循环里的if-else写法：

bool isAlphanumeric(char &c) {
    if(c >= 'A' && c <= 'Z') c += 32;
    else if(c >= '0' && c <= '9' || c >= 'a' && c <= 'z') return true;
    else return false;
    return true;
}

bool isPalindrome(string s) {
    int i = 0, j = s.length()-1;
    while(i < j) {
        if(!isAlphanumeric(s[i])) ++i;
        else if(!isAlphanumeric(s[j])) --j;
        else if(s[i++] != s[j--]) return false;
    }

    return true;
}

　python3

class Solution:
    def isPalindrome(self, s: str) -> bool:
        if len(s) == 0 or len(s) == 1:
            return True
        i = 0
        j = len(s) - 1
        validChar = 'abcdefghijklmnopqrstuvwxyz0123456789'
        while True:
            while i < len(s) and s[i].lower() not in validChar:
                i += 1
            while j >= 0 and s[j].lower() not in validChar:
                j -= 1
            if i >= j:
                return True
            if s[i].lower() != s[j].lower(): 
                return False
            i += 1
            j -= 1