一、题目

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

“…zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd…”.
Given a string s, return the number of unique non-empty substrings of s are present in base.

Example 1:

Input: s = “a”
Output: 1
Explanation: Only the substring “a” of s is in base.
Example 2:

Input: s = “cac”
Output: 2
Explanation: There are two substrings (“a”, “c”) of s in base.
Example 3:

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

Constraints:

1 <= s.length <= 105
s consists of lowercase English letters.

二、题解

class Solution {
public:int findSubstringInWraproundString(string s) {int n = s.size();vector<int> a(n,0);for(int i = 0;i < n;i++){a[i] = s[i] - 'a';}//dp[0]代表s中以a结尾的子串，最大延伸长度是多少(根据base串规则)vector<int> dp(26,0);dp[a[0]] = 1;int len = 1;for(int i = 1;i < n;i++){int cur = a[i],pre = a[i-1];if((pre == 25 && cur == 0) || pre == cur - 1) len++;else len = 1;dp[cur] = max(dp[cur],len);}int res = 0;for(int i = 0;i < 26;i++){res += dp[i];}return res;}
};