Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:
 Input: "abc"
 Output: 3
 Explanation: Three palindromic strings: "a", "b", "c".
 Example 2:
 Input: "aaa"
 Output: 6
 Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
 Note:
 The input string length won't exceed 1000.
 好久没刷Leetcode,今天心血来潮搞一道题目，首先用最直观的方法解决这个问题，解法思路为先从两个字母的开始判断，然后判断三个字母的，然后依次累加直到判断整个字符串。这种解法中存在冗余，比如网站中的提示部分：
 If “aba” is a palindrome, is “xabax” and palindrome? Similarly is “xabay” a palindrome?
 如果aba是回文，在aba前后增加相同的字母依然还是回文，如果在aba前后增加不同的字符则不是回文。 

int isPalindromic(char* s,int sSize)
{
    for(int i=0;i < sSize/2;i++)
    {
        if(s[i] != s[sSize-1-i])
        {
            return 0;
        }
    }
    
    return 1;
}
int countSubstrings(char* s) 
{
    int slen = strlen(s);
     /*单个字符为回文*/
     int ret = slen;
    for(int i= 2;i <= slen;i++)
    {
        for(int j = 0;j< slen-i+1;j++)
        {
            if(isPalindromic(&s[j],i))
            {
                ret++;
            }
        }
    }
    return ret;
}


int main()
{
    char s[]="abcdeeff";
    printf("%d\n",countSubstrings(s));
    
    return 0;
}

本方法用时60ms，击败34.62%的提交。
 下面时一种击败100% 用时4ms的解决实现： 

int function(char* s,int count)
{
    int ret=0;
    
    for(int i=1;i<=count;i++)
    {
        if(s[count-i]==s[count+i])
        {
            ret++;
        }
        else
        {
            break;
        }
    }
    
    int temp=count;
    for(int i = 1;count >= 0;i++)
    {
        if(s[count] == s[temp + i])
        {
            ret++;
        }
        else
        {
            break;
        }
        count--;
    }
    return ret;
}
int countSubstrings(char* s) 
{
    int slen = strlen(s);
    /*单个字符为回文*/
    int ret = slen;
    for(int i = 0;i < slen;i++)
    {
       ret= ret + function(s,i);
    }
    
    return ret;
}