To find a subset of a string, for example, to find a substring of 1234, there are 1, 2, 3, 4, 12, 13, 14, 23, 24, 34, 123, 124, 134, 234, 1234.. At that time, there was no I figured out how to do it. When I entered the second side, the interviewer asked the question of the written test. Now I think I should do it. Unfortunately, I did not think carefully after the written test. I learned this lesson and referred to some ideas on the Internet. Using recursion is the easiest way to find all subsets of a string.
All elements in the set have two possibilities for each subset: in the subset or not.
These two possibilities for individual elements combine to form all subsets of a set. Each set has 2^n subsets. The proper subset is: (2^n)-1
We put the element in the state of being in the subset or not in the boolean array.
/**
* 求1234的子串，例如：1，2，3，4，12，13，14，23，24，34，123，124，134，234，1234
*
* 集合中的所有元素对于每一个子集来说，都有两种可能性：在子集中或是不在子集中。
* 各个元素的这两种可能性组合起来，组成了一个集合的所有子集。每一个集合都有2^n个子集。
*
* 我们把元素这种在或不在子集中状态，放置在boolean数组中。
* @author c
*
*/
public class Test {
private static void get_Subset(String str, int start, int end, boolean[] b) {
// TODO Auto-generated method stub
if(start==end){//当start==end时，即集合中的所有元素都已经在或者不在该子集中，输出该子集后，return跳出该层递归。
int i = 0;
for(;i<end;i++){
if(b[i]){
System.out.print(str.charAt(i));
}
}
System.out.print(" ");
return ;
}else{
b[start] = false;//代表数组中索引为start的元素不在该子集中
get_Subset(str, start+1, end, b);//而后进入递归，元素索引向后加一，同样索引为start+1的元素也有在或者不在该子集中两种可能性
b[start] = true;
get_Subset(str, start+1, end, b);
}
}
public static void main(String[] args) {
String str = "1234";
//我们把元素这种在或不在子集中状态，放置在boolean数组中。
boolean []b = new boolean[str.length()];
get_Subset(str,0,str.length(),b);
}
}