问实现模糊搜索建议/词完成

EN

/**
 * <ul>
 * <b><i>searchDistance</i></b><br>
 * <br>
 * <code>&nbsp;public static int searchDistance(String key, String match)</code><br>
 * <br>
 * Gets the Levenshtein distance between <code>key</code> and <code>match</code>. <br>
 * If <code>useAsterisk</code> is true, then the follwing applies: If <code>key</code> is shorter than <code>match</code>, the asterisk <code>'*'</code> is appended to it until the lengths are equal. Asterisks can be used in <code>key</code> to signify 'any character.'
 * @param key - The text to search for
 * @param match - The text to compare <code>key</code> against
 * @param useAsterisk - Whether or not to use asterisks for the purpose described above
 * @return the Levenshtein distance between <code>key</code> and <code>match</code>.
 *         </ul>
 */
public static int searchDistance(String key, String match, boolean useAsterisk) {
    while (key.length() < match.length()) {
        key = key + "*";
    }

    int[][] matrix = new int[key.length() + 1][match.length() + 1];

    for (int i = 0; i < matrix.length; i++) {
        matrix[i][0] = i;
    }

    for (int i = 0; i < matrix[0].length; i++) {
        matrix[0][i] = i;
    }

    for (int a = 1; a < matrix.length; a++) {
        for (int b = 1; b < matrix[0].length; b++) {
            matrix[a][b] = Math.min(Math.min(matrix[a - 1][b] + 1, matrix[a][b - 1] + 1), matrix[a - 1][b - 1] + (key.charAt(a - 1) == match.charAt(b - 1) || key.charAt(a - 1) == '*' ? 0 : 1));
        }
    }

    return matrix[matrix.length - 1][matrix[0].length - 1];
}