Leetcode_valid-palindrome

最新推荐文章于 2020-05-17 12:09:08 发布

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本文介绍了一种算法，用于判断字符串是否为回文，仅考虑字母数字字符并忽略大小写。通过过滤非字母数字字符，并利用STL反转功能进行比较实现。提供了两种C++实现方案。

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地址： http://oj.leetcode.com/problems/valid-palindrome/

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.

思路：只把大小写字母和数字留下，用stl反置一下再对比是否相同。

另外有个函数可以检测是否是字母和数字。不过在这题里不是很合适，因为有大小写的缘故。

int isalnum ( int c )

参考代码：

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.empty())
            return true;
        string str;
        for(int i = 0; i <s.length(); ++i)
        {
            if((s[i]>='a'&&s[i]<='z') || (s[i]>='0'&&s[i]<='9'))
                str+=s[i];
            else if(s[i]>='A'&&s[i]<='Z')
                str+=s[i]-'A'+'a';
        }
        string rev_str = str;
        reverse(rev_str.begin(), rev_str.end());
        return rev_str == str;
    }
};

  
   //SECOND TRIAL
  
  
   class 
   Solution 
   {
  
  
   public
   :
  
  
       
   bool 
   isPalindrome
   (
   string 
   s
   ) 
   {
  
  
           
   if
   (
   s
   .
   empty
   ())
  
  
               
   return 
   true
   ;
  
  
           
   int 
   cnt 
   = 
   0
   ;
  
  
           
   for
   (
   int 
   i 
   = 
   0
   ; 
   i
   <
   s
   .
   length
   (); 
   ++
   i
   )
  
  
           
   {
  
  
               
   if
   ((
   s
   [
   i
   ]
   >=
   '0' 
   && 
   s
   [
   i
   ]
   <=
   '9'
   )
   ||
   (
   s
   [
   i
   ]
   >=
   'a' 
   &&
   s
   [
   i
   ]
   <=
   'z'
   ))
  
  
                   
   s
   [
   cnt
   ++
   ] 
   = 
   s
   [
   i
   ];
  
  
               
   else 
   if
   (
   s
   [
   i
   ]
   >=
   'A' 
   && 
   s
   [
   i
   ]
   <=
   'Z'
   )
  
  
                   
   s
   [
   cnt
   ++
   ] 
   = 
   s
   [
   i
   ] 
   + 
   'a' 
   - 
   'A'
   ;
  
  
           
   }
  
  
           
   s 
   = 
   s
   .
   substr
   (
   0
   , 
   cnt
   );
  
  
           
   string 
   revs
   (
   s
   .
   rbegin
   (), 
   s
   .
   rend
   ());
  
  
           
   return 
   revs
   ==
   s
   ;
  
  
       
   }
  
  
   };