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## Problem

We define the string s to be the infinite wraparound string of “abcdefghijklmnopqrstuvwxyz”, so s will look like this:

• “…zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd….”.

Given a string p, return the number of unique non-empty substrings of p are present in s.

Example 1:

Input: p = "a"
Output: 1
Explanation: Only the substring "a" of p is in s.


Example 2:

Input: p = "cac"
Output: 2
Explanation: There are two substrings ("a", "c") of p in s.


Example 3:

Input: p = "zab"
Output: 6
Explanation: There are six substrings ("z", "a", "b", "za", "ab", and "zab") of p in s.


Constraints:

• 1 <= p.length <= 105
• p consists of lowercase English letters.

## Code

class Solution {
public int findSubstringInWraproundString(String p) {
int[] dict = new int[26];
int len = 0;

for(int i = 0; i < p.length(); i++){
char curr = p.charAt(i);

if(i > 0 && (curr - p.charAt(i - 1) == 1 || (p.charAt(i - 1) == 'z' && curr == 'a'))){
len++;
} else {
len = 1;
}

dict[curr - 'a'] = Math.max(dict[curr - 'a'], len);
}

int res = 0;
for(int num : dict){
res += num;
}

return res;
}
}