Valid Palindrome leetcode java

題目：

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.

 

題解：

 這道題的幾個點，

一就是alphanumeric characters and ignoring cases，字母和數字，忽略大小寫。

二就是考慮空字符串是否為回文，最好在面試時候問下面試官，這里是認為空字符串是回文。

因為忽略大小寫，所以就統一為大寫。

然后就判斷當前檢查字符是否符合范圍，否則大小指針挪動。

如果發現有大小指針指向的值有不同的，就返回false，否則，繼續檢查。

最后返回true。

代碼如下：

 
           1         
          public  
          static  
          boolean isPalindrome(String s) { 
          
 
           2              
          if(s.length()==0) 
          
 
           3                  
          return  
          true; 
          
 
           4              
          
 
           5             s = s.toUpperCase(); 
          
 
           6              
          int low1 = 'A', high1 = 'Z'; 
          
 
           7              
          int low2 = '0', high2 = '9'; 
          
 
           8              
          int low = 0, high = s.length()-1; 
          
 
           9              
          
 
          10              
          while(low < high){ 
          
 
          11                  
          if((s.charAt(low)<low1||s.charAt(low)>high1) 
          
 
          12                     && (s.charAt(low)<low2||s.charAt(low)>high2)){ 
          
 
          13                         low++; 
          
 
          14                          
          continue; 
          
 
          15                     } 
          
 
          16                      
          
 
          17                  
          if((s.charAt(high)<low1||s.charAt(high)>high1) 
          
 
          18                     && (s.charAt(high)<low2||s.charAt(high)>high2)){ 
          
 
          19                         high--; 
          
 
          20                          
          continue; 
          
 
          21                     } 
          
 
          22                  
          if(s.charAt(low) == s.charAt(high)){ 
          
 
          23                     low++; 
          
 
          24                     high--; 
          
 
          25                 } 
          else 
          
 
          26                      
          return  
          false; 
          
 
          27             } 
          
 
          28              
          return  
          true; 
          
 
          29         }