467. Unique Substrings in Wraparound String

题目：

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 strin

翻译：

考虑到字符串的字符串是无限的概括“有”，所以会看起来像这样：“…zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd的…”。
现在我们有另一个字符串P。你的工作是找出许多独特的非空的P在美国特别是现在，你输入的字符串P和你需要输出不同的非空的子串的字符串S，P数
注：P仅由小写英文字母组成，P的大小可能超过10000。
例1：
输入：“A”
输出：1
说明：只有子”“字符串”“在字符串s
例2：
输入：“CAC”
输出：2
说明：有两个子字符串“A”、“C”“CAC”字符串在字符串s
例3：
输入：“朱达”
输出：6
说明：有六子“Z”，“A”、“B”、“杂”、“AB”、“朱达”字符串“朱达”的字符串

解题代码：

#include<iostream>
#include<string>
using namespace std;


class Solution {
public:
    int findSubstringInWraproundString(string p) {
        vector<int> letters(26, 0);
        int res = 0, len = 0;
        for (int i = 0; i < p.size(); i++) {
            int cur = p[i] - 'a';
            if (i > 0 && p[i - 1] != (cur + 26 - 1) % 26 + 'a') len = 0;
            if (++len > letters[cur]) {
                res += len - letters[cur];
                letters[cur] = len;
            }
        }
        return res;
    }
};

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