《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：

 1 // 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.

 2 #include <iostream>

 3 #include <string>

 4 #include <unordered_set>

 5 #include <vector>

 6 using namespace std;

 7 

 8 class Solution {

 9 public:

10     int KMPMatch(const string &word, const string &pattern) {

11         int index;

12         int pos;

13         int result;

14         

15         lw = word.length();

16         lp = pattern.length();

17         calculateNext(pattern);

18 

19         index = pos = 0;

20         result = 0;

21         while (index < lw) {

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

23                 ++index;

24                 ++pos;

25             } else {

26                 pos = next[pos];

27             }

28             

29             if (pos == lp) {

30                 pos = 0;

31                 ++result;

32             }

33         }

34         

35         return result;

36     };

37     

38     ~Solution() {

39         next.clear();

40     };

41 private:

42     int lw;

43     int lp;

44     vector<int> next;

45     

46     void calculateNext(const string &pattern) {

47         int i = 0;

48         int j = -1;

49         

50         next.resize(lp + 1);

51         next[0] = -1;

52         while (i < lp) {

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

54                 ++i;

55                 ++j;

56                 next[i] = j;

57             } else {

58                 j = next[j];

59             }

60         }

61     };

62 };

63 

64 int main()

65 {

66     string text;

67     unordered_set<string> dict;

68     Solution sol;

69     int n, i;

70     int sum;

71     

72     while (cin >> n && n > 0) {

73         for (i = 0; i < n; ++i) {

74             cin >> text;

75             dict.insert(text);

76         }

77         cin >> text;

78         

79         sum = 0;

80         unordered_set<string>::const_iterator usit;

81         for (usit = dict.begin(); usit != dict.end(); ++usit) {

82             sum += sol.KMPMatch(text, *usit);

83         }

84         cout << sum << endl;

85         

86         dict.clear();

87     }

88     

89     return 0;

90 }