125. Valid Palindrome

125. Valid Palindrome

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• Difficulty: Easy
• Contributors: Admin

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.

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import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Solution {
    public boolean isPalindrome(String s) {
      boolean flag=false;
if(s==""||"".equals(s.trim()))
{
flag=true;
}
//过滤字符 

String regEx="[`~!@#$%^&*()+=|{}':;',\\[\\].<>/?~！@#￥%……&*（）——+|{}【】‘；：”“’。，、？]"; 
Pattern p = Pattern.compile(regEx); 
Matcher m = p.matcher(s);
  String num=m.replaceAll("").trim();
  String test="";//结果存放
  char[] old=num.toCharArray();//转换为字符
  //去掉空格
  for(int i=0;i<old.length;i++){
  if(old[i]==' '||" ".equals(old[i])){
  continue;
  }
  else{
  test=test+old[i];
  }
  }
 
     //根据字符串创建一个字符缓存类对象sb
       StringBuffer sb=new StringBuffer(test);
      //将字符缓存中的内容倒置
       sb.reverse();
      //计算出str与sb中对应位置字符相同的个数n
       int n=0;
       test=test.toLowerCase();
       String sb2=sb.toString().toLowerCase();
      
       
       for(int i=0;i<test.length();i++){
        if(test.charAt(i)==sb2.charAt(i) )
         n++;
       }
      //如果所有字符都相等，即n的值等于str的长度，则str就是回文。
          if(n==test.length()){
          flag=true;
          }
return flag;  
    }
}