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Use one-dimensional vector.

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chriseth 2018-02-09 17:53:30 +01:00 committed by Alex Beregszaszi
parent 12c3eb8880
commit 5747985e6a
1 changed files with 8 additions and 7 deletions
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15
libdevcore/StringUtils.cpp
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@ -48,7 +48,8 @@ size_t dev::stringDistance(string const& _str1, string const& _str2)
size_t n1 = _str1.size(); size_t n1 = _str1.size();
size_t n2 = _str2.size(); size_t n2 = _str2.size();
// Optimize by storing only last 2 rows and current row. So first index is considered modulo 3 // Optimize by storing only last 2 rows and current row. So first index is considered modulo 3
vector<vector<size_t>> dp(3, vector<size_t>(n2 + 1)); // This is a two-dimensional array of size 3 x (n2 + 1).
vector<size_t> dp(3 * (n2 + 1));
// In this dp formulation of Damerau–Levenshtein distance we are assuming that the strings are 1-based to make base case storage easier. // In this dp formulation of Damerau–Levenshtein distance we are assuming that the strings are 1-based to make base case storage easier.
// So index accesser to _name1 and _name2 have to be adjusted accordingly // So index accesser to _name1 and _name2 have to be adjusted accordingly
@ -60,9 +61,9 @@ size_t dev::stringDistance(string const& _str1, string const& _str2)
x = max(i1, i2); x = max(i1, i2);
else else
{ {
size_t left = dp[(i1-1) % 3][i2]; size_t left = dp[(i1 - 1) % 3 + i2 * 3];
size_t up = dp[i1 % 3][i2-1]; size_t up = dp[(i1 % 3) + (i2 - 1) * 3];
size_t upleft = dp[(i1 - 1) % 3][i2-1]; size_t upleft = dp[((i1 - 1) % 3) + (i2 - 1) * 3];
// deletion and insertion // deletion and insertion
x = min(left + 1, up + 1); x = min(left + 1, up + 1);
if (_str1[i1-1] == _str2[i2-1]) if (_str1[i1-1] == _str2[i2-1])
@ -74,12 +75,12 @@ size_t dev::stringDistance(string const& _str1, string const& _str2)
// transposing // transposing
if (i1 > 1 && i2 > 1 && _str1[i1 - 1] == _str2[i2 - 2] && _str1[i1 - 2] == _str2[i2 - 1]) if (i1 > 1 && i2 > 1 && _str1[i1 - 1] == _str2[i2 - 2] && _str1[i1 - 2] == _str2[i2 - 1])
x = min(x, dp[(i1 - 2) % 3][i2 - 2] + 1); x = min(x, dp[((i1 - 2) % 3) + (i2 - 2) * 3] + 1);
} }
dp[i1 % 3][i2] = x; dp[(i1 % 3) + i2 * 3] = x;
} }
return dp[n1 % 3][n2]; return dp[(n1 % 3) + n2 * 3];
} }
string dev::quotedAlternativesList(vector<string> const& suggestions) string dev::quotedAlternativesList(vector<string> const& suggestions)