algorithm - 迭代版本的Damerau–Levenshtein距离

Levenshtein距离可以通过以下方式使用两行迭代计算：
https://en.wikipedia.org/wiki/Levenshtein_distance#Iterative_with_two_matrix_rows
我遇到了考虑到换位的Optimal String alignment distance。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，我无法将该页上伪代码算法的扩展移植到迭代版本的代码中任何帮助都非常感谢。

最佳答案

你需要三行来计算这个新版本，我无法检查代码，但我对它很有信心：

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wikipedia实现。

关于algorithm - 迭代版本的Damerau–Levenshtein距离，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/39332845/