题目描述：
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.找出给定字符串的子字符串，满足它们是"...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...."的子字符串，解决这道题的方法是分别确定以各个字母为结尾的最长子字符串，例如以‘a’为结尾的子字符串的最大程度为k，那么一共有k个以‘a’为结尾的子字符串满足题意。
class Solution {
public:
int findSubstringInWraproundString(string p) {
vector<int> count(26,0);
int len=1;
for(int i=0;i<p.size();i++)
{
if(i>0&&(p[i]-p[i-1]==1||p[i-1]-p[i]==25)) len++;
else len=1;
count[p[i]-'a']=max(count[p[i]-'a'],len);
}
int result=0;
for(int i=0;i<26;i++) result+=count[i];
return result;
}
};