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July 26, 2016 23:06
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longest common substring - C++ solution, using Dynamic programmng - code source: http://blog.iderzheng.com/longest-common-substring-problem-optimization/
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int longestCommonSubstring_n2_n2(const string& str1, const string& str2) | |
{ | |
size_t size1 = str1.size(); | |
size_t size2 = str2.size(); | |
if (size1 == 0 || size2 == 0) return 0; | |
vector<vector<int> > table(size1, vector<int>(size2, 0)); | |
// the start position of substring in original string | |
int start1 = -1; | |
int start2 = -1; | |
// the longest length of common substring | |
int longest = 0; | |
// record how many comparisons the solution did; | |
// it can be used to know which algorithm is better | |
int comparisons = 0; | |
for (int j = 0; j < size2; ++j) | |
{ | |
++comparisons; | |
table[0][j] = (str1[0] == str2[j] ? 1 :0); | |
} | |
for (int i = 1; i < size1; ++i) | |
{ | |
++comparisons; | |
table[i][0] = (str1[i] == str2[0] ? 1 :0); | |
for (int j = 1; j < size2; ++j) | |
{ | |
++comparisons; | |
if (str1[i] == str2[j]) | |
{ | |
table[i][j] = table[i-1][j-1]+1; | |
} | |
} | |
} | |
for (int i = 0; i < size1; ++i) | |
{ | |
for (int j = 0; j < size2; ++j) | |
{ | |
if (longest < table[i][j]) | |
{ | |
longest = table[i][j]; | |
start1 = i-longest+1; | |
start2 = j-longest+1; | |
} | |
} | |
} | |
#ifdef IDER_DEBUG | |
cout<< "(first, second, comparisions) = (" | |
<< start1 << ", " << start2 << ", " << comparisons | |
<< ")" << endl; | |
#endif | |
return longest; | |
} |
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