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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