Valid Palindrome

Valid Palindrome

问题：

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

思路：

　　简单的数学推导

我的代码:

public class Solution {

    public boolean isPalindrome(String s) {

        if(s == null || s.length() == 0)  return true;

        int left = 0;

        int right = s.length() - 1;

        while(left < right)

        {

            while(!isAlpha(s.charAt(left)) && left < right) left++;

            while(!isAlpha(s.charAt(right)) && left < right) right--;

            if(!isEqual(s.charAt(left), s.charAt(right))) return false;

            left ++;

            right --;

        }

        return true;

    }

    public boolean isAlpha(char c)

    {

        return ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'));

    }

    public boolean isEqual(char left, char right)

    {

       return (left == right || Math.abs(left-right) == 32);

    }

}

View Code

他人代码：

public class Solution {

    public boolean isPalindrome(String s) {

        if (s == null || s.length() == 0) {

            return true;

        }



        int front = 0;

        int end = s.length() - 1;

        while (front < end) {

            while (front < s.length() && !isvalid(s.charAt(front))){ // nead to check range of a/b

                front++;

            }



            if (front == s.length()) { // for emtpy string “.,,,”     

                return true; 

            }           



            while (end >= 0 && ! isvalid(s.charAt(end))) { // same here, need to check border of a,b

                end--;

            }



            if (Character.toLowerCase(s.charAt(front)) != Character.toLowerCase(s.charAt(end))) {

                break;

            } else {

                front++;

                end--;

            }

        }



        return end <= front; 

    }



    private boolean isvalid (char c) {

        return Character.isLetter(c) || Character.isDigit(c);

    }

}

View Code

学习之处：

• Character.isLetter(c) Character.isDigit(c) Character.toLowerCase(c)  Character还有这几个API不错
• 写代码的时候，尽量的简洁和功能划分，比如我的代码里面的isAlpha和isEqual都可以用一行代码解决，这样就不要用多行，注意功能划分。