algorithm - Damerau–Levenshtein 距离的迭代版本

Question

Levenshtein 距离可以使用两行以这种方式迭代计算:

https://en.wikipedia.org/wiki/Levenshtein_distance#Iterative_with_two_matrix_rows

我遇到了 Optimal String alignment distance这确实考虑到了换位。维基百科说它可以使用常规 Levenshtein 算法的直接扩展来计算:

if i > 1 and j > 1 and a[i-1] = b[j-2] and a[i-2] = b[j-1] then
    d[i, j] := minimum(d[i, j],
                       d[i-2, j-2] + cost)  // transposition

但是，我无法将该页面上的伪代码算法扩展移植到迭代版本的代码中。非常感谢任何帮助。

Answer 1

你需要三行来计算这个新版本，我无法检查代码但我对此很有信心:

int DamerauLevenshteinDistance(string s, string t)
{
// degenerate cases
if (s == t) return 0;
if (s.Length == 0) return t.Length;
if (t.Length == 0) return s.Length;

// create two work vectors of integer distances
int[] v0 = new int[t.Length + 1];
int[] v1 = new int[t.Length + 1];
int[] v2 = new int[t.Length + 1];

// initialize v0 (the previous row of distances)
// this row is A[0][i]: edit distance for an empty s
// the distance is just the number of characters to delete from t
for (int i = 0; i < v0.Length; i++)
    v0[i] = i;

    // compute v1

    v1[0] = 0;

    // use formula to fill in the rest of the row
    for (int j = 0; j < t.Length; j++)
    {
        var cost = (s[0] == t[j]) ? 0 : 1;
        v1[j + 1] = Minimum(v1[j] + 1, v0[j + 1] + 1, v0[j] + cost);
    }

if (s.Length == 1) {
    return v1[t.Length];
}

for (int i = 1; i < s.Length; i++)
{
    // calculate v2 (current row distances) from the previous rows v0 and v1

    // first element of v2 is A[i+1][0]
    //   edit distance is delete (i+1) chars from s to match empty t
    v2[0] = i + 1;

    // use formula to fill in the rest of the row
    for (int j = 0; j < t.Length; j++)
    {
        var cost = (s[i] == t[j]) ? 0 : 1;
        v2[j + 1] = Minimum(v2[j] + 1, v1[j + 1] + 1, v1[j] + cost);
        if (j > 0 && s[i] = t[j-1] && s[i-1] = t[j])
             v2[j + 1] = Minimum(v2[j+1],
                   v0[j-1] + cost);
    }

    // copy v2 (current row) to v1 (previous row) and v1 to v0 for next iteration
    for (int j = 0; j < v0.Length; j++)
        v0[j] = v1[j];
        v1[j] = v2[j];
}

return v2[t.Length];
} 

原始代码来自上面提到的维基百科实现。