125. Valid Palindrome

题目：

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.

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

题解：

根据题意用两个指针对冲法。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isPalindrome(String s) {
        if(s == null)
            return true;
        int lo = 0, hi = s.length() - 1;
        s = s.toLowerCase();
        
        while(lo <= hi) {
            if(!isAlphanumeric(s.charAt(lo))) {
                lo++;
                continue;
            }
            if(!isAlphanumeric(s.charAt(hi))) {
                hi--;
                continue;
            }
            if(s.charAt(lo) != s.charAt(hi))
                return false;
            else {
                lo++;
                hi--;
            }
        }
        
        return true;
    }
    
    private boolean isAlphanumeric(char c) {
        if(Character.isDigit(c) || Character.isLetter(c))
            return true;
        else
            return false;
    }
}

二刷：

方法和一刷一样，双指针向中间夹逼，忽略非alphanumeric value，

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) {
            return false;
        }
        s = s.toLowerCase();
        int lo = 0, hi = s.length() - 1;
        while (lo <= hi) {
            while (lo < hi && !isDigitOrAlphabet(s.charAt(lo))) {
                lo++;
            }    
            while (lo < hi && !isDigitOrAlphabet(s.charAt(hi))) {
                hi--;
            }
            if (s.charAt(lo) == s.charAt(hi)) {
                lo++;
                hi--;
            } else {
                return false;
            }
        }
        return true;
    }
    
    private boolean isDigitOrAlphabet(char c) {
        if ((c <= '9' && c >= '0') || (c <= 'z' && c >= 'a')) {
            return true;
        }
        return false;
    }
    
}

使用库方法的

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) {
            return false;
        }
        int lo = 0, hi = s.length() - 1;
        while (lo <= hi) {
            while (lo < hi && !Character.isLetterOrDigit(s.charAt(lo))) {
                lo++;
            }    
            while (lo < hi && !Character.isLetterOrDigit(s.charAt(hi))) {
                hi--;
            }
            if (Character.toLowerCase(s.charAt(lo)) == Character.toLowerCase(s.charAt(hi))) {
                lo++;
                hi--;
            } else {
                return false;
            }
        }
        return true;
    }
}

三刷:

Java:

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) return false;
        int lo = 0, hi = s.length() - 1;
        while (lo <= hi) {
            char loChar = s.charAt(lo);
            if (!Character.isLetterOrDigit(loChar)) {
                lo++;
                continue;
            }
            char hiChar = s.charAt(hi);
            if (!Character.isLetterOrDigit(hiChar)) {
                hi--;
                continue;
            }
            if (Character.toLowerCase(loChar) != Character.toLowerCase(hiChar)) return false;
            lo++;
            hi--;
        }
        return true;
    }
}

Reference:

https://leetcode.com/discuss/23989/accepted-pretty-java-solution-271ms