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Merge LLVM r242069, make spell checking allocate less memory.

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Nico Weber 2015-09-07 14:26:19 -07:00
parent 2810ffb2b5
commit fdb885dec7
1 changed files with 20 additions and 18 deletions
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38
src/edit_distance.cc
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@ -28,40 +28,42 @@ int EditDistance(const StringPiece& s1,
// http://en.wikipedia.org/wiki/Levenshtein_distance
//
// Although the algorithm is typically described using an m x n
// array, only two rows are used at a time, so this implementation
// just keeps two separate vectors for those two rows.
// array, only one row plus one element are used at a time, so this
// implementation just keeps one vector for the row. To update one entry,
// only the entries to the left, top, and top-left are needed. The left
// entry is in row[x-1], the top entry is what's in row[x] from the last
// iteration, and the top-left entry is stored in previous.
int m = s1.len_;
int n = s2.len_;
vector<int> previous(n + 1);
vector<int> current(n + 1);
for (int i = 0; i <= n; ++i)
previous[i] = i;
vector<int> row(n + 1);
for (int i = 1; i <= n; ++i)
row[i] = i;
for (int y = 1; y <= m; ++y) {
current[0] = y;
int best_this_row = current[0];
row[0] = y;
int best_this_row = row[0];
int previous = y - 1;
for (int x = 1; x <= n; ++x) {
int old_row = row[x];
if (allow_replacements) {
current[x] = min(previous[x-1] + (s1.str_[y-1] == s2.str_[x-1] ? 0 : 1),
min(current[x-1], previous[x])+1);
row[x] = min(previous + (s1.str_[y - 1] == s2.str_[x - 1] ? 0 : 1),
min(row[x - 1], row[x]) + 1);
}
else {
if (s1.str_[y-1] == s2.str_[x-1])
current[x] = previous[x-1];
if (s1.str_[y - 1] == s2.str_[x - 1])
row[x] = previous;
else
current[x] = min(current[x-1], previous[x]) + 1;
row[x] = min(row[x - 1], row[x]) + 1;
}
best_this_row = min(best_this_row, current[x]);
previous = old_row;
best_this_row = min(best_this_row, row[x]);
}
if (max_edit_distance && best_this_row > max_edit_distance)
return max_edit_distance + 1;
current.swap(previous);
}
return previous[n];
return row[n];
}