leetcode.5. Longest Palindromic Substring

字符串遍历，从中间分出奇数回文和偶数回文两种情况不断更新长度。我一开陷入入误区把字符串分为奇偶结果偶数的可以过，基数的abb过不了。所以就是分析回文就好

char* longestPalindrome(char* s)
    {
        int len = strlen(s);
        if (len <= 1)
        { return s; }
        int start = 0;
        int maxlen = 0;
        for (int i = 1; i < len; i++)
        {
            int low = i - 1;
            int high = i;
            while (low >= 0 && high < len && s[low] == s[high])
            {
                low--;
                high++;
            }
            if (high - low - 1 > maxlen)
            {
                maxlen = high - low - 1;
                start = low + 1;
            }
            low = i - 1; high = i + 1;
            while (low >= 0 && high < len && s[low] == s[high])
            {
                low--;
                high++;
            }
            if (high - low - 1 > maxlen)
            {
                maxlen = high - low - 1;
                start = low + 1;
            }
        }
        char *arr = (char *)malloc(sizeof(char) * maxlen+1000);
        int i = 0;
        for (; i < maxlen; i++)
        {
            arr[i] = s[start++];
        }
        arr[i] = '\0';
        return arr;
    }