你怎么实现Levenshtein distance在德尔菲?
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09-06-2019 - |
题
我张贴这在精神回答你自己的问题。
这个问题我曾是:我如何可以实现的Levenshtein算法计算的编辑之间的距离两个字符串 这里描述的, 在德尔菲?
只是一个注意性能:这件事是非常快。在我的桌面上(2.33Ghz的双重核心,2GB ram、WinXP),我可以跑过一系列100K串在不到一秒钟。
解决方案
function EditDistance(s, t: string): integer;
var
d : array of array of integer;
i,j,cost : integer;
begin
{
Compute the edit-distance between two strings.
Algorithm and description may be found at either of these two links:
http://en.wikipedia.org/wiki/Levenshtein_distance
http://www.google.com/search?q=Levenshtein+distance
}
//initialize our cost array
SetLength(d,Length(s)+1);
for i := Low(d) to High(d) do begin
SetLength(d[i],Length(t)+1);
end;
for i := Low(d) to High(d) do begin
d[i,0] := i;
for j := Low(d[i]) to High(d[i]) do begin
d[0,j] := j;
end;
end;
//store our costs in a 2-d grid
for i := Low(d)+1 to High(d) do begin
for j := Low(d[i])+1 to High(d[i]) do begin
if s[i] = t[j] then begin
cost := 0;
end
else begin
cost := 1;
end;
//to use "Min", add "Math" to your uses clause!
d[i,j] := Min(Min(
d[i-1,j]+1, //deletion
d[i,j-1]+1), //insertion
d[i-1,j-1]+cost //substitution
);
end; //for j
end; //for i
//now that we've stored the costs, return the final one
Result := d[Length(s),Length(t)];
//dynamic arrays are reference counted.
//no need to deallocate them
end;
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