《Cracking the Coding Interview》——第18章：难题——题目8

2014-04-29 03:10

题目：给定一个长字符串S和一个词典T，进行多模式匹配，统计S中T单词出现的总个数。

解法：这是要考察面试者能不能写个AC自动机吗？对面试题来说太难了吧？我不会，所以只写了个KMP用N次的方法。

代码：

 // 18.8 Given a list of words and a piece of text, find out how many times in total, all words appear in the text.
 #include <iostream>
 #include <string>
 #include <unordered_set>
 #include <vector>
 using namespace std;

 class Solution {
 public:
     int KMPMatch(const string &word, const string &pattern) {
         int index;
         int pos;
         int result;

         lw = word.length();
         lp = pattern.length();
         calculateNext(pattern);

         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;
     };

     ~Solution() {
         next.clear();
     };
 private:
     int lw;
     int lp;
     vector<int> next;

     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 main()
 {
     string text;
     unordered_set<string> dict;
     Solution sol;
     int n, i;
     int sum;

     while (cin >> n && n > ) {
         for (i = ; i < n; ++i) {
             cin >> text;
             dict.insert(text);
         }
         cin >> text;

         sum = ;
         unordered_set<string>::const_iterator usit;
         for (usit = dict.begin(); usit != dict.end(); ++usit) {
             sum += sol.KMPMatch(text, *usit);
         }
         cout << sum << endl;

         dict.clear();
     }

     return ;
 }