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.

判读一个字符串是否是回文字符串(只判断期中的字符和数字)。

1、两个栈,从前向后和从后向前,然后判断两个栈是否相同。

public class Solution {
public boolean isPalindrome(String s) {
Stack stack1 = new Stack<Character>();
Stack stack2 = new Stack<Character>();
s = s.toLowerCase();
char[] word = s.toCharArray();
int len = s.length();
for( int i = 0;i<len;i++){
if( (word[i] >= '0' && word[i] <= '9') || (word[i]>='a' && word[i]<='z') )
stack1.push(word[i]);
if( (word[len-1-i] >= '0' && word[len-1-i] <= '9') || (word[len-1-i]>='a' && word[len-1-i]<='z') )
stack2.push(word[len-1-i]);
} return stack1.equals(stack2); }
}

2、直接用两个指针记录left和right即可。

public class Solution {
public boolean isPalindrome(String s) { char[] word = s.toLowerCase().toCharArray();
int len = s.length();
int left = 0,right = len-1;
while( left < right ){ if( !((word[left] >= '0' && word[left] <= '9') || (word[left]>='a' && word[left]<='z' )) )
left++;
else if( !((word[right] >= '0' && word[right] <= '9') || (word[right]>='a' && word[right]<='z' )) )
right--;
else if( word[left] == word[right]){
left++;
right--;
}else
return false;
}
return true; }
}

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