双指针

28. Implement strStr()

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

125. Valid Palindrome 有效回文串

判断一个字符串是否是回文串，除字母数字外其他的字符不考虑，不考虑大小写的区别

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.

使用双指针，注意事项是大写字母的ASCII比小写字母小

另外这里不区分大小写字母, 运行不通过

class Solution {
public:
    bool isPalindrome(string s) {
        if(!s.size()) return true;
        int n=s.size();
        int first=0,last=n-1;
        while(first<last)
        {
            while(first<last&&(s[first]<'A'||s[first]>'z')&&(s[first]<'0'||s[first]>'9')) first++;
            while(first<last&&(s[last]<'A'||s[last]>'z')&&(s[last]<'0'||s[last]>'9')) last--;
            if(first<last&&s[first]!=s[last])
            {
              if(s[first]>='A'&&s[first]<'z'&&s[last]>='A'&&s[last]<='z')
              {
                 if(s[first]!=s[last]+'a'-'A'&&s[first]!=s[last]+'A'-'a')
                      return false; 
                  else break;
              }
                else return false; 
            }
                    first++;
                    last--;
            }
         return true;
        }
       
    
};

简易：使用两个库函数

class Solution {
public:
    bool isPalindrome(string s) {
         int n=s.size();
        int i=0,j=n-1;
        while(i<j)
        {
            while(i<j&&isalnum(s[i])==false) i++;
            while(i<j&&isalnum(s[j])==false) j--;
            if(toupper(s[i])!=toupper(s[j])) return false;
            i++;
            j--;
        }
        return true;
        }
       
    
};