1.8 Assume you have a method isSubstring which checks if one word is a substring of another. Given two strings, s1 and s2, write code to check if s2 is a rotation of si using only one call to isSubstring (e.g.,"waterbottle"is a rotation of "erbottlewat").
这道题给定两个字符串,让我们判断其中一个是否是另一个的旋转字符串,并给了我们一个例子来说明旋转字符串,比如waterbottle是erbottlewat的旋转字符串。而且还给我们了一个isSubstring函数可以调用,这个函数是用来判断一个字符串是否是另一个字符串的子字符串,不过规定了我们只能调用一次。这题要用到一个小技巧,就是加入我们将s1重复两次,变成s1s1,那么s2如果是s1s1的子字符串,那么它们就互为旋转字符串,就拿题目中的梯子来分析:
若令 x = wat y = erbottle
则 s1 = xy s2 = yx
若令 s1s1 = xyxy
则 s2 一定是 s1s1的子字符串
class Solution {public: bool isRotation(string s1, string s2) { if (s1.size() != s2.size() || s1.empty()) return false; string s1s1 = s1 + s1; return isSubstring(s1s1, s2); }};