LeetCode--Valid Palindrome

Posted 2020-11-10 SkyeAngel

Valid Palindrome   

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

题目中是对数字和字母进行回文判断，而对其他的字符 ! ? . , " ; : @ # % $ * - / 等都是忽略的，所以有个问题，就是要忽略掉这些乱七八糟的东西，而一般的加条件进行判断也不太好用，

看到讨论区有 用Character.isLetterOrDigit(ch)来进行筛选，很厉害了。

 

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++){
            if(Character.isLetterOrDigit(s.charAt(i))){
                sb.append(s.charAt(i));
            }
        }
        String str = sb.toString().toLowerCase();
        String res = sb.reverse().toString().toLowerCase();
        if(str.equals(res)) return true;
        return false;
    }
}

　　

那句话可以这样代替，很笨

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++){
            if((s.charAt(i) >= ‘a‘ && s.charAt(i) <= ‘z‘) || (s.charAt(i) >= ‘A‘ && s.charAt(i) <= ‘Z‘) || (s.charAt(i) >= ‘0‘ && s.charAt(i) <= ‘9‘) ){
                sb.append(s.charAt(i));
            }
        }
        String str = sb.toString().toLowerCase();
        String res = sb.reverse().toString().toLowerCase();
        if(str.equals(res)) return true;
        return false;
    }
}