Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.

Note: p consists of only lowercase English letters and the size of p might be over 10000.

Example 1:

Input: "a"

Output: 1

Explanation: Only the substring "a" of string "a" is in the string �s.

Example 2:

Input: "cac"

Output: 2

Explanation: There are two substrings "a", "c" of string "cac" in the string s.

Example 3:

Input: "zab"

Output: 6

Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

    public int findSubstringInWraproundString(String p) {
            int[] count = new int[26];
            int maxlengthcur =0;
            int N = p.length();
            for(int i=0;i<N;i++) {
                if(i>0&&(p.charAt(i)-p.charAt(i-1)==1||(p.charAt(i)-p.charAt(i-1)==-25))) {
                    maxlengthcur++;
                }else {
                    maxlengthcur=1;
                }
                int t = p.charAt(i)-'a';
                count[t]=Math.max(count[t], maxlengthcur);
            }
            int sum=0;
            for(int i=0;i<26;i++) {
                sum = sum+count[i];
            }
        return sum;
}

这其实是一个动态规划的问题；

举例， ABCD BCD CD D。 ABCD会囊括（BCD，CD，D）这些字符串的子字符串。