Select three characters to form a password.

Given the a-z0-9, in which select three characters to form a password, output all the circumstances, the program implementation [java] import java. util. arrayList; import java. util. list; public class Demo {static char [] buf = {'1', '2', '3', '4 '}; static List <String> list = new ArrayList <String> (); public static void main (String [] args) {select (buf, list, 3 ); www.2cto.com for (String str: list) {System. out. println (str);} System. out. println ("In total:" + list. size ();} public s Tatic void select (char [] source, List <String> arrayList, int num) {int l = source. length; char [] temp = new char [num]; System. arraycopy (source, 0, temp, 0, num); arrayList. add (new String (temp); for (int I = num; I <l; I ++) {for (int j = 0; j <num; j ++) {char tempChar = temp [j]; temp [j] = source [I]; arrayList. add (new String (temp); temp [j] = tempChar ;}}2. Many programs use strings in large quantities. For different strings, we hope to be able to determine their similar programs. We define a set of operation methods to make the two strings different from each other the same. The specific operation method is: 1. modify a character (for example, replace "a" with "B"); 2. add a character (for example, change "abdd" to "aebdd"); 3. delete A character (for example, change "traveling" to "traveling"). For example, for the strings "abcdefg" and "abcdef, we believe that the goal can be achieved by adding/decreasing a "g. The preceding two solutions only need to be used once. The number of times required for this operation is defined as the distance between two strings, and the similarity is equal to the reciprocal of "distance + 1. That is to say, the distance between "abcdefg" and "abcdef" is 1, and the similarity is 1/2 = 0.5. Given any two strings, can you write an algorithm to calculate their similarity? [Java] public class Levenshtein {public int ld (String str1, String str2) {// Distance int [] [] d; int n = str1.length (); int m = str2.length (); int I; // iterate str1 int j; // iterate str2 char limit; // str1 char ch2; // str2 int temp; 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;} f Or (I = 1; I <= n; I ++) {character = str1.charAt (I-1); // match str2 for (j = 1; j <= m; j ++) {ch2 = str2.charAt (j-1); if (temperature = ch2) {temp = 0;} else {temp = 1 ;} d [I] [j] = min (d [I-1] [j] + 1, d [I] [j-1] + 1, d [I-1] [j-1] + temp) ;}} return d [n] [m];} /*** get the smallest number **/private int min (int one, int two, int three) {return (one = one <two? One: two) <three? One: three;} www.2cto.com/*** obtain the similarity between two strings */public float getSimilarityRatio (String str, String target) {return (1/(float) ld (str, target) + 1);} public static void main (String [] args) {Levenshtein lt = new Levenshtein (); String str = "huang "; string target = "huang"; System. out. println (lt. getSimilarityRatio (str, target ));}}