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  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：

 // http://www.careercup.com/question?id=5188169901277184
 #include <iostream>
 #include <string>
 #include <vector>
 using namespace std;

 class Solution {
 public:
     int countWord(const string &word, string &pattern) {
         lw = (int)word.length();
         lp = (int)pattern.length();

         if (lw ==  || lp == ) {
             return ;
         }

         if (lw < lp) {
             return ;
         }

         int result = ;
         if (isPalindrome(pattern)) {
             calculateNext(pattern);
             result += KMPMatch(word, pattern);
         } else {
             calculateNext(pattern);
             result += KMPMatch(word, pattern);
             reverse(pattern.begin(), pattern.end());
             calculateNext(pattern);
             result += KMPMatch(word, pattern);
             reverse(pattern.begin(), pattern.end());
             result *= ;
         }
         next.clear();

         return result;
     }
 private:
     int lw;
     int lp;
     vector<int> next;

     bool isPalindrome(const string &s) {
         int len = (int)s.length();
         int i;

         if (len <= ) {
             return true;
         }

         for (i = ; i < len -  - i; ++i) {
             if (s[i] != s[len -  - i]) {
                 return false;
             }
         }

         return true;
     }

     void calculateNext(const string &pattern) {
         int i = ;
         int j = -;

         next.resize(lp + );
         next[] = -;
         while (i < lp) {
             if (j == - || pattern[i] == pattern[j]) {
                 ++i;
                 ++j;
                 next[i] = j;
             } else {
                 j = next[j];
             }
         }
     }

     int KMPMatch(const string &word, const string &pattern)
     {
         int index;
         int pos;
         int result;

         index = pos = ;
         result = ;
         while (index < lw) {
             if (pos == - || word[index] == pattern[pos]) {
                 ++index;
                 ++pos;
             } else {
                 pos = next[pos];
             }

             if (pos == lp) {
                 pos = ;
                 ++result;
             }
         }

         return result;
     }
 };

 int main()
 {
     string word, pattern;
     Solution sol;

     while (cin >> word >> pattern) {
         cout << sol.countWord(word, pattern) << endl;
     }

     return ;
 }