《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 = 0;
        result = 0;
        while (index < lw) {
            if (pos == -1 || word[index] == pattern[pos]) {
                ++index;
                ++pos;
            } else {
                pos = next[pos];
            }
            
            if (pos == lp) {
                pos = 0;
                ++result;
            }
        }
        
        return result;
    };
    
    ~Solution() {
        next.clear();
    };
private:
    int lw;
    int lp;
    vector<int> next;
    
    void calculateNext(const string &pattern) {
        int i = 0;
        int j = -1;
        
        next.resize(lp + 1);
        next[0] = -1;
        while (i < lp) {
            if (j == -1 || 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 > 0) {
        for (i = 0; i < n; ++i) {
            cin >> text;
            dict.insert(text);
        }
        cin >> text;
        
        sum = 0;
        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 0;
}