原文:
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 s1 using only one call to isSubstring ( i.e., “waterbottle” is a rotation of “erbottlewat”).
译文:
假设你有一个isSubstring函数,可以检测一个字符串是否是另一个字符串的子串。 给出字符串s1和s2,只使用一次isSubstring就能判断s2是否是s1的旋转字符串, 请写出代码。旋转字符串:"waterbottle"是"erbottlewat"的旋转字符串。
解答:
如果s1是s2的旋转字符串,那么我们一定能找到一个切割点将s1分割成xy,并且yx等于s2。例如:
s1 = xy = waterbottle x = wat y = erbottle s2 = yx = erbottlewat
那么我们可以知道yx一定是xyxy的一个子串,也就是说,s2一定是s1s1的子串。所以这个题目的解法也就一目了然了。
public class Main { public static boolean isSubstring(String s1, String s2) { if(s1 == null || s2 == null) return false; if(s1.contains(s2)) return true; else return false; } public static boolean isRotation(String s1, String s2) { return isSubstring(s1+s1,s2); } public static void main(String args[]) { String s1 = "waterbottle"; String s2 = "erbottlewat"; System.out.println(isRotation(s1,s2)); } }