Levenshtein distance 编辑距离

Posted 2020-08-17 风云力

编辑距离，又称Levenshtein距离，是指两个字串之间，由一个转成另一个所需的最少编辑操作次数。许可的编辑操作包括将一个字符替换成另一个字符，插入一个字符，删除一个字符

 

实现方案：

1. 找出最长公共子串长度

 

 

参考代码：

apache commons-lang

public static int getLevenshteinDistance(CharSequence s, CharSequence t) {
    if (s == null || t == null) {
        throw new IllegalArgumentException("Strings must not be null");
    }

    /*
       The difference between this impl. and the previous is that, rather
       than creating and retaining a matrix of size s.length() + 1 by t.length() + 1,
       we maintain two single-dimensional arrays of length s.length() + 1.  The first, d,
       is the ‘current working‘ distance array that maintains the newest distance cost
       counts as we iterate through the characters of String s.  Each time we increment
       the index of String t we are comparing, d is copied to p, the second int[].  Doing so
       allows us to retain the previous cost counts as required by the algorithm (taking
       the minimum of the cost count to the left, up one, and diagonally up and to the left
       of the current cost count being calculated).  (Note that the arrays aren‘t really
       copied anymore, just switched...this is clearly much better than cloning an array
       or doing a System.arraycopy() each time  through the outer loop.)

       Effectively, the difference between the two implementations is this one does not
       cause an out of memory condition when calculating the LD over two very large strings.
     */

    int n = s.length(); // length of s
    int m = t.length(); // length of t

    if (n == 0) {
        return m;
    } else if (m == 0) {
        return n;
    }

    if (n > m) {
        // swap the input strings to consume less memory
        final CharSequence tmp = s;
        s = t;
        t = tmp;
        n = m;
        m = t.length();
    }

    int p[] = new int[n + 1]; //‘previous‘ cost array, horizontally
    int d[] = new int[n + 1]; // cost array, horizontally
    int _d[]; //placeholder to assist in swapping p and d

    // indexes into strings s and t
    int i; // iterates through s
    int j; // iterates through t

    char t_j; // jth character of t

    int cost; // cost

    for (i = 0; i <= n; i++) {
        p[i] = i;
    }

    for (j = 1; j <= m; j++) {
        t_j = t.charAt(j - 1);
        d[0] = j;

        for (i = 1; i <= n; i++) {
            cost = s.charAt(i - 1) == t_j ? 0 : 1;
            // minimum of cell to the left+1, to the top+1, diagonally left and up +cost
            d[i] = Math.min(Math.min(d[i - 1] + 1, p[i] + 1), p[i - 1] + cost);
        }

        // copy current distance counts to ‘previous row‘ distance counts
        _d = p;
        p = d;
        d = _d;
    }

    // our last action in the above loop was to switch d and p, so p now
    // actually has the most recent cost counts
    return p[n];
}