我坐在这里,用Java为我的主程序编写了一些算法(好吧,到目前为止是第一个)。我编写的levenshtein算法很好,这要归功于wiki对新手的伪代码和一个很好的教程:D
然后我决定升级到Damerau,并添加了额外的行,但后来我读到它不是DL algo,而是OptimalStringAlignmentDistance。我试着阅读actionscript代码,以了解我还需要添加什么才能将其添加到DL中,但结果却被搞糊涂了。我去过不同的地方,代码看起来与Java相似,但它们也都使用了错误的伪代码。
在花了半天时间后,我放弃了,决定在这里提问。有没有人可以帮我把这段代码升级到Java中的Damerau-Levenshtein?
public class LevensteinDistance {
private static int Minimum(int a, int b, int c) {
return Math.min(Math.min(a, b), c);
}
private static int Minimum (int a, int b) {
return Math.min(a, b);
}
public static int computeLevensteinDistance(String s, String t){
int d[][];
int n; // length of s
int m; // length of t
int i; // iterates through s
int j; // iterates through t
char s_i; // ith character of s
char t_j; // jth character of t
int cost; // cost
n = s.length ();
m = t.length ();
if (n == 0) {
return m;
}
if (m == 0) {
return n;
}
d = new int[n+1][m+1];
for (i = 0; i <= n; i++) {
d[i][0] = i;
}
for (j = 0; j <= m; j++) {
d[0][j] = j;
}
for(i = 1; i <= n; i++) {
s_i = s.charAt (i - 1);
for(j = 1; j <= m; j++) {
t_j = t.charAt (j - 1);
if(s_i == t_j){
cost = 0;
}else{
cost = 1;
}
d[i][j] = Minimum(d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1] + cost);
if(i > 1 && j > 1 && s_i == t_j-1 && s_i-1 == t_j){
d[i][j] = Minimum(d[i][j], d[i-2][j-2] + cost);
}
}
}
return d[n][m];
}
// public static void main(String[] args0){
// String a = "I decided it was best to ask the forum if I was doing it right";
// String b = "I thought I should ask the forum if I was doing it right";
// System.out.println(computeLevensteinDistance(a, b));
// }
}这是Damerau–Levenshtein distance algorithm的维基百科页面
发布于 2011-05-18 02:36:48
您的问题在于在条件中引用字符串中的前几个字符。在你的原始代码中,你有:
if(i > 1 && j > 1 && s_i == t_j-1 && s_i-1 == t_j){
d[i][j] = Minimum(d[i][j], d[i-2][j-2] + cost);
}问题是t_j-1和s_i-1的值。它们说的是第i个字符s和t减1,其中算法说你想要第i个减1的字符。例如,如果字符串s是"AFW“,而i是1,那么
s_i - 1 = E; //the character value (s[1]='F') minus 1 = 'E'
s.charAt(i-1) = A; //i-1 = 0, s[0] = 'A'所以你的条件应该是:
if(i > 1 && j > 1 && s_i == t.charAt(j-1) && s.charAt(i-1) == t_j) {
d[i][j] = Minimum(d[i][j], d[i-2][j-2] + cost);
}编辑:不幸的是,通过阅读代码,我不能理解算法,但是这里是来自维基百科页面的ActionScript示例的翻译,它给出了一个与您的示例匹配的输出:
public static int damerauLevenshteinDistance(
String a, String b, int alphabetLength) {
final int INFINITY = a.length() + b.length();
int[][] H = new int[a.length()+2][b.length()+2];
H[0][0] = INFINITY;
for(int i = 0; i<=a.length(); i++) {
H[i+1][1] = i;
H[i+1][0] = INFINITY;
}
for(int j = 0; j<=b.length(); j++) {
H[1][j+1] = j;
H[0][j+1] = INFINITY;
}
int[] DA = new int[alphabetLength];
Arrays.fill(DA, 0);
for(int i = 1; i<=a.length(); i++) {
int DB = 0;
for(int j = 1; j<=b.length(); j++) {
int i1 = DA[b.charAt(j-1)];
int j1 = DB;
int d = ((a.charAt(i-1)==b.charAt(j-1))?0:1);
if(d==0) DB = j;
H[i+1][j+1] =
min(H[i][j]+d,
H[i+1][j] + 1,
H[i][j+1]+1,
H[i1][j1] + (i-i1-1) + 1 + (j-j1-1));
}
DA[a.charAt(i-1)] = i;
}
return H[a.length()+1][b.length()+1];
}
private static int min(int ... nums) {
int min = Integer.MAX_VALUE;
for (int num : nums) {
min = Math.min(min, num);
}
return min;
}发布于 2014-03-21 09:46:24
我认为SparseArray可以用于DA,这样就不需要知道字母表的确切大小。
SparseArray<Integer> DA = new SparseArray<Integer>();
...
int i1 = DA.get(b.charAt(j - 1), 0);https://stackoverflow.com/questions/6033631
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