Careercup - Microsoft面试题 - 5188169901277184

2014-05-12 06:12

题目链接

原题：

Write a function to retrieve the number of a occurrences of a substring(even the reverse of a substring) in a string without using the java substring() method. 



Ex: 'dc' in 'abcd' occurs 2 times (dc, cd).

题目：写一个函数来统计模式串在文本中出现的次数，不过这次模式的反转也算数。比如“abcd”中dc算出现两次。

解法：按照这种算法，应该是原模式串匹配一次，反转模式串再匹配一次，然后结果加起来乘以二。如果模式串本身是回文串的话，那就只匹配一次。匹配使用KMP算法。

代码：

  1 // http://www.careercup.com/question?id=5188169901277184

  2 #include <iostream>

  3 #include <string>

  4 #include <vector>

  5 using namespace std;

  6 

  7 class Solution {

  8 public:

  9     int countWord(const string &word, string &pattern) {

 10         lw = (int)word.length();

 11         lp = (int)pattern.length();

 12         

 13         if (lw == 0 || lp == 0) {

 14             return 0;

 15         }

 16         

 17         if (lw < lp) {

 18             return 0;

 19         }

 20         

 21         int result = 0;

 22         if (isPalindrome(pattern)) {

 23             calculateNext(pattern);

 24             result += KMPMatch(word, pattern);

 25         } else {

 26             calculateNext(pattern);

 27             result += KMPMatch(word, pattern);

 28             reverse(pattern.begin(), pattern.end());

 29             calculateNext(pattern);

 30             result += KMPMatch(word, pattern);

 31             reverse(pattern.begin(), pattern.end());

 32             result *= 2;

 33         }

 34         next.clear();

 35         

 36         return result;

 37     }

 38 private:

 39     int lw;

 40     int lp;

 41     vector<int> next;

 42     

 43     bool isPalindrome(const string &s) {

 44         int len = (int)s.length();

 45         int i;

 46 

 47         if (len <= 1) {

 48             return true;

 49         }

 50 

 51         for (i = 0; i < len - 1 - i; ++i) {

 52             if (s[i] != s[len - 1 - i]) {

 53                 return false;

 54             }

 55         }

 56 

 57         return true;

 58     }

 59 

 60     void calculateNext(const string &pattern) {

 61         int i = 0;

 62         int j = -1;

 63         

 64         next.resize(lp + 1);

 65         next[0] = -1;

 66         while (i < lp) {

 67             if (j == -1 || pattern[i] == pattern[j]) {

 68                 ++i;

 69                 ++j;

 70                 next[i] = j;

 71             } else {

 72                 j = next[j];

 73             }

 74         }

 75     }

 76     

 77     int KMPMatch(const string &word, const string &pattern)

 78     {

 79         int index;

 80         int pos;

 81         int result;

 82         

 83         index = pos = 0;

 84         result = 0;

 85         while (index < lw) {

 86             if (pos == -1 || word[index] == pattern[pos]) {

 87                 ++index;

 88                 ++pos;

 89             } else {

 90                 pos = next[pos];

 91             }

 92             

 93             if (pos == lp) {

 94                 pos = 0;

 95                 ++result;

 96             }

 97         }

 98         

 99         return result;

100     }

101 };

102 

103 int main()

104 {

105     string word, pattern;

106     Solution sol;

107     

108     while (cin >> word >> pattern) {

109         cout << sol.countWord(word, pattern) << endl;

110     }

111     

112     return 0;

113 }