[LeetCode] 125. Valid Palindrome Java

Posted 2020-10-12 BrookLearnData

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.

题意及分析：给出一个字符串，只考虑其中的字母或者数字字符，问字符串是否是一个回文，注意空字符串也是回文。

方法一：先遍历一次字符串去除非数字字母字符，然后判断是否是回文，复杂度为n+n/2

class Solution {
    public boolean isPalindrome(String s) {
        if(s == null || s.length() == 0 ) return true;

        StringBuilder sb = new StringBuilder();
        for(int i=0;i<s.length();i++){      //取出标点符号，只留下字母
            char ch = s.charAt(i);
            if(Character.isLetterOrDigit(ch)){
                sb.append(ch);
            }
        }

        //全部变为小写字母
        String newStr = sb.toString().toLowerCase();
        for(int i=0;i<newStr.length()/2;i++){
            if(newStr.charAt(i)!=newStr.charAt(newStr.length()-1-i))
                return false;
        }
        return true;
    }
}

方法二：直接判断，每次分别从前和从后找到一个字母数字字符，若不相等直接返回false，复杂度最多为n

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