Careercup

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：

// 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 == 0 || lp == 0) {
            return 0;
        }
        
        if (lw < lp) {
            return 0;
        }
        
        int result = 0;
        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 *= 2;
        }
        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 <= 1) {
            return true;
        }

        for (i = 0; i < len - 1 - i; ++i) {
            if (s[i] != s[len - 1 - i]) {
                return false;
            }
        }

        return true;
    }

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

int main()
{
    string word, pattern;
    Solution sol;
    
    while (cin >> word >> pattern) {
        cout << sol.countWord(word, pattern) << endl;
    }
    
    return 0;
}