125. Valid Palindrome

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.

链接： http://leetcode.com/problems/valid-palindrome/

一刷，用两个指针从两个方向，出现的问题

1. string methods掌握不熟，str.isalpha(), str.isdigit(), str.isalnum()

2. 内部两个while loop之后需要再检查index是否valid，否则或者应该跳到外循环检查

3. 不要忘记更新下标

class Solution(object):
    def isPalindrome(self, s):
        if not s or s == '':
            return True
        i, j = 0, len(s) - 1
        
        while i <= j:
            while i < len(s) and not s[i].isalnum():
                i += 1
            while j >= 0 and not s[j].isalnum():
                j -= 1
            if i == len(s) or j < 0:
                break
            if s[i].isalpha() and s[j].isalpha() and s[i].lower() == s[j].lower() or s[i].isdigit() and s[j].isdigit() and s[i] == s[j]:
                pass
            else:
                return False
            i += 1
            j -= 1
        return True

对比相同方法别人的代码，可以有以下改动：

class Solution(object):
    def isPalindrome(self, s):
        i, j = 0, len(s) - 1
        while i <= j:
            while i < j and not s[i].isalnum():
                i += 1
            while i < j and not s[j].isalnum():
                j -= 1
            if s[i].lower() != s[j].lower():
                return False
            i += 1
            j -= 1
        return True

别人还有更pythonic但是需要额外空间复杂度的做法，很赞，平时工作完全能写得出，但是刷题时就忘记了：

class Solution(object):
    def isPalindrome(self, s):
        cleanlist = [c for c in s.lower() if c.isalnum()]
        return cleanlist == cleanlist[::-1]

2/18/2017, Java

注意Java Character各种函数

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) return true;
        int start = 0, end = s.length() - 1;
        char c1, c2;
        boolean isDigitS, isLetterS, isDigitE, isLetterE;
        while (start <= end) {
            c1 = s.charAt(start);
            isDigitS = Character.isDigit(c1);
            isLetterS = Character.isLetter(c1);
            if (!isDigitS && !isLetterS) {
                start++;
                continue;
            }
            c2 = s.charAt(end);
            isDigitE = Character.isDigit(c2);
            isLetterE = Character.isLetter(c2);
            if (!isDigitE && !isLetterE) {
                end--;
                continue;
            }
            if (isDigitE && isDigitS && c1 == c2 || isLetterS && isLetterE && Character.toLowerCase(c1) == Character.toLowerCase(c2)) {
                start++;
                end--;
                continue;
            }
            return false;
        }
        return true;
    }
}