详见:https://leetcode.com/problems/unique-substrings-in-wraparound-string/description/ C++: class Solution { public: int findSubstringInWraproundString(string p) { vector<int> cnt(26, 0); int len = 0; for (int i = 0; i < p.size(); ++i) { if (i >…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/unique-substrings-in-wraparound-string/description/ 题目描述: Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so…

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…

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…

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…

467. 环绕字符串中唯一的子字符串 把字符串 s 看作是"abcdefghijklmnopqrstuvwxyz"的无限环绕字符串,所以 s 看起来是这样的:"-zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd-". 现在我们有了另一个字符串 p .你需要的是找出 s 中有多少个唯一的 p 的非空子串,尤其是当你的输入是字符串 p ,你需要输出字符串 s 中 p 的不同的非空子串的数目. 注意: p…

环绕字符串中的唯一子字符串 把字符串 s 看作是"abcdefghijklmnopqrstuvwxyz"的无限环绕字符串,所以 s 看起来是这样的:"...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". 现在我们有了另一个字符串 p .你需要的是找出 s 中有多少个唯一的 p 的非空子串,尤其是当你的输入是字符串 p ,你需要输出字符串 s 中 p 的不同的非空子串的数目. 注意: p …

python判断字符串中是否包含子字符串 s = '1234问沃尔沃434' if s.find('沃尔沃') != -1:     print('存在') else:     print('不存在')…

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…