使用完全符合您要求的 Levenshtein 距离的变体:
输出
-MKNLASREVNIYVNGKLV
QM---ASREVNIYVNGKL-
代码:
public class Main {
public static void main(String[] args) {
String[] aligned = align("MKNLASREVNIYVNGKLV", "QMASREVNIYVNGKL");
System.out.println(aligned[0]);
System.out.println(aligned[1]);
}
public static String[] align(String a, String b) {
int[][] T = new int[a.length() + 1][b.length() + 1];
for (int i = 0; i <= a.length(); i++)
T[i][0] = i;
for (int i = 0; i <= b.length(); i++)
T[0][i] = i;
for (int i = 1; i <= a.length(); i++) {
for (int j = 1; j <= b.length(); j++) {
if (a.charAt(i - 1) == b.charAt(j - 1))
T[i][j] = T[i - 1][j - 1];
else
T[i][j] = Math.min(T[i - 1][j], T[i][j - 1]) + 1;
}
}
StringBuilder aa = new StringBuilder(), bb = new StringBuilder();
for (int i = a.length(), j = b.length(); i > 0 || j > 0; ) {
if (i > 0 && T[i][j] == T[i - 1][j] + 1) {
aa.append(a.charAt(--i));
bb.append("-");
} else if (j > 0 && T[i][j] == T[i][j - 1] + 1) {
bb.append(b.charAt(--j));
aa.append("-");
} else if (i > 0 && j > 0 && T[i][j] == T[i - 1][j - 1]) {
aa.append(a.charAt(--i));
bb.append(b.charAt(--j));
}
}
return new String[]{aa.reverse().toString(), bb.reverse().toString()};
}
}