《Cracking the Coding Interview》——第1章：数组和字符串——题目8

2014-03-18 02:12

题目：判断一个字符串是否由另一个字符串循环移位而成。

解法：首先长度必须相等。然后将第一个串连拼两次，判断第二个串是否在这个连接串中。

代码：

 1 // 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 s1 using only one call to isSubstring (i.e., “waterbottle” is a rotation of “erbottlewat”).

 2 #include <cstdio>

 3 #include <cstring>

 4 using namespace std;

 5 

 6 class Solution {

 7 public:

 8     bool isStringRotation(char *s1, char *s2) {

 9         if (s1 == nullptr || s2 == nullptr) {

10             return false;

11         }

12         

13         int len1, len2;

14         

15         len1 = strlen(s1);

16         len2 = strlen(s2);

17         if (len1 != len2) {

18             return false;

19         }

20         

21         const int MAXLEN = 1005;

22         static char tmp[MAXLEN];

23         

24         tmp[0] = 0;

25         strcat(tmp, s1);

26         strcat(tmp, s1);

27         return strstr(tmp, s2) != nullptr;

28     }

29 private:

30     bool isSubstring(char *haystack, char *needle) {

31         if (haystack == nullptr || needle == nullptr) {

32             return false;

33         }

34         

35         return strstr(haystack, needle) != nullptr;

36     }

37 };

38 

39 int main()

40 {

41     char s1[1005];

42     char s2[1005];

43     Solution sol;

44     

45     while (scanf("%s%s", s1, s2) == 2) {

46         printf("\"%s\" is ", s2);

47         if (!sol.isStringRotation(s1, s2)) {

48             printf("not ");

49         }

50         printf("a rotation of \"%s\".\n", s1);

51     }

52     

53     return 0;

54 }