Difficulty:easy

More:【目錄】LeetCode Java實現

Description

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

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

Intuition

Use two pointers, one points to head (walk toward tail), another points to tail(walk toward head), if two characters are different, then return false.

Be careful: There are no different between a lower case letter and its upper case in this problem.

Solution

With the help of API:

    public boolean isPalindrome(String s) {
        if(s==null || s.length()<0)
            return false;
        int head=0;
        int tail=s.length()-1;
        while(head<=tail){
            if(!Character.isLetterOrDigit(s.charAt(head)))
                head++;
            else if(!Character.isLetterOrDigit(s.charAt(tail)))
                tail--;
            else{
                if(Character.toLowerCase(s.charAt(head))!=Character.toLowerCase(s.charAt(tail)))
                    return false;
                head++;
                tail--;
            }
        }
        return true;
    }

Without API:

    public boolean isPalindrome(String s) {
        if(s==null)
            return false;       
        int i=0;
        int j=s.length()-1;
        
        while(i<=j){
            while(i<s.length() && (!(isAlpha(s.charAt(i)) || isNumeric(s.charAt(i)))))
                i++;
            while(j>=0 && (!(isAlpha(s.charAt(j)) || isNumeric(s.charAt(j)))))
                j--;
            if(i<s.length() && j>=0 && !isSame(s.charAt(i),s.charAt(j)) )
                return false;
            i++;
            j--;
        }
        return true;
    }
    
    private boolean isAlpha(char c){
        if( (c>=‘a‘ && c<=‘z‘) || (c>=‘A‘ && c<=‘Z‘))
            return true;
        return false;
    }

    private boolean isNumeric(char c){
        if(c>=‘0‘ && c<=‘9‘)
            return true;
        return false;
    }

    private boolean isSame(char a,char b){
        if(a==b)
            return true;
        int len=‘A‘-‘a‘;
        if((isAlpha(a)&&isAlpha(b)) && (a-b==len || b-a==len))
            return true;
        return false;
    }
    

Complexity

Time complexity : O(n)

Space complexity : O(1)

What I‘ve learned

1. API: Character.isLetterOrDigit(c), Character.toLowerCase(c), Character.toUpperCase(c)

2. DO NOT forget head++ and tail-- in the line 14 and line 15

3. DO NOT forget i<s.length() && j>=0 in the lines 8,10,12

4. s.length() not s.length

More:【目錄】LeetCode Java實現

【LeetCode】125. Valid Palindrome