125. Valid Palindrome

題目：

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.

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

一刷，用兩個指針從兩個方向，出現的問題

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

2. 內部兩個while loop以後須要再檢查index是否valid，不然或者應該跳到外循環檢查

3. 不要忘記更新下標

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

對比相同方法別人的代碼，能夠有如下改動：

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

別人還有更pythonic可是須要額外空間複雜度的作法，很贊，平時工做徹底能寫得出，可是刷題時就忘記了：

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各類函數

 1 public class Solution {
 2     public boolean isPalindrome(String s) {
 3         if (s == null) return true;
 4         int start = 0, end = s.length() - 1;
 5         char c1, c2;
 6         boolean isDigitS, isLetterS, isDigitE, isLetterE;
 7         while (start <= end) {
 8             c1 = s.charAt(start);
 9             isDigitS = Character.isDigit(c1);
10             isLetterS = Character.isLetter(c1);
11             if (!isDigitS && !isLetterS) {
12                 start++;
13                 continue;
14             }
15             c2 = s.charAt(end);
16             isDigitE = Character.isDigit(c2);
17             isLetterE = Character.isLetter(c2);
18             if (!isDigitE && !isLetterE) {
19                 end--;
20                 continue;
21             }
22             if (isDigitE && isDigitS && c1 == c2 || isLetterS && isLetterE && Character.toLowerCase(c1) == Character.toLowerCase(c2)) {
23                 start++;
24                 end--;
25                 continue;
26             }
27             return false;
28         }
29         return true;
30     }
31 }