Levenshtein distance 編輯距離

編輯距離，又稱Levenshtein距離，是指兩個字串之間，由一個轉成另外一個所需的最少編輯操做次數。許可的編輯操做包括將一個字符替換成另外一個字符，插入一個字符，刪除一個字符

 

實現方案：

1. 找出最長公共子串長度

 

 

參考代碼：

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];}