Question

我对Damerau-Levenshtein编辑距离有问题（操作：删除，卖淫，插入，交换）

我需要使用Damerau-Levenshtein矩阵来具体改变第一个带有已使用操作列表的String

示例：

str1: pair
str2: pari
----
distance: 1
SWAP [2] "i": pair -> pari

我的代码可以正确使用矩阵生成

 public int[][] matrixgen(String str1, String str2, int rows, int cols)
                return LevenshteinDistanceMatrix

我在第二部分遇到了麻烦，事实上在某些情况下这个算法并没有以正确的顺序执行操作

public int editDistDP(String str1, String str2, int rows, int cols) {

    SString sstr_1 = new SString(str1);
    SString sstr_2 = new SString(str2);

    int i = rows;
    int j = cols;

    int[][] dp = matrixgen(sstr_1.to_String(), sstr_2.to_String(), rows, cols);

    while (i > 0 && j > 0) {

        int min;
        System.out.println(sstr_1.to_String() + " -> " + sstr_2.to_String());


        if (sstr_1.equals_To(sstr_2)) break;

        if (i > 1 && j > 1) {
            min = min_v2(dp[i][j - 1], 
                    dp[i - 1][j],  
                    dp[i - 1][j - 1], 
                    dp[i - 2][j - 2]); //minumum as 4
        } else {
            min = min(dp[i][j - 1], 
                    dp[i - 1][j],  
                    dp[i - 1][j - 1]); //minimun as 3
        }

        if (min == dp[i][j]) {
            i--;
            j--;
        } 
        else if (min == dp[i - 1][j]) {
            System.out.print("DELETE [" + i + "]: " + sstr_1.to_String().charAt(i - 1));
            sstr_1.delete(i); //edit sstr_1 and delete in posix i
            i--;
        } 
        else if (min == dp[i][j - 1]) {

            System.out.print("INSERT [" + i + "]: " + sstr_2.to_String().charAt(j - 1));
            sstr_1.insert(sstr_2.to_String().charAt(j - 1), i);  //edit sstr_1 and insert sstr_2.to_String().charAt(j - 1) in sstr_1 into posix i           
            j--;
        } 
        else if (min == dp[i - 1][j - 1]) {

            System.out.print("REPLACE [" + i + "]: " + sstr_2.to_String().charAt(j - 1));
            sstr_1.replace(i, sstr_2.to_String().charAt(j - 1)); //edit sstr_1 and replace sstr_2.to_String().charAt(j - 1) in sstr_1's posix i       
            i--;
            j--;
        } 
        else if (min == dp[i - 2][j - 2]) {
            System.out.print("SWAP [" + (i - 2) + "]: " + sstr_1.to_String().charAt(i - 2));
            sstr_1.swap(i - 2); //edit sstr_1 and swap sstr_1's char in posix i-2 with sstr_1's char in posix (i-2)+1
            i -= 1;
            j -= 1;
        }
    }

    System.out.println(sstr_1.to_String() + " == " + sstr_2.to_String());

    return dp[rows][cols];
}

在“对”的情况下 - ＆gt; “pari”，此算法有效

在“volubile”的情况下 - ＆gt; “olubiel”，距离为2，我们有交换+删除这是输出：

START: volubile -> olubiel
SWAP [6]: "l" >> volubiel | olubiel //correct
volubiel -> olubiel
REPLACE [7]: "e" >> volubiee | olubiel //ERROR: it has to delete "v" not replace
volubiee -> olubiel

然后返回0

使用Damerau-Levenshtein编辑距离来修改字符串

0 个答案: