Merge LLVM r242069, make spell checking allocate less memory.

1 files changed, 20 insertions, 18 deletions

diff --git a/src/edit_distance.cc b/src/edit_distance.cc
index a6719d3..3bb62b8 100644
--- a/src/edit_distance.cc
+++ b/src/edit_distance.cc
@@ -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];
 }