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;
}
}