原创转载请注明出处：http://agilestyle.iteye.com/blog/2360659
找出一个Set集合中的所有子集
比如全集为 {1, 2, 3}
子集则有{}，{1}，{2}，{3}，{1，2}，{1，3}，{2，3}，{1，2，3}
此算法的核心思想：递归
package org.fool.java.collections;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
public class GetAllSubSetTest {
public static void main(String[] args) {
Set<Integer> set = new HashSet<>();
set.add(1);
set.add(2);
set.add(3);
set.add(4);
// set to array
Integer[] nums = set.toArray(new Integer[set.size()]);
// use lambda to transform Integer[] to int[]
printSubSet(Arrays.stream(nums).mapToInt(i -> i).toArray());
}
// step 1: decide how many elements in a sub-set to be printed
public static void printSubSet(int[] nums) {
for (int i = 0; i <= nums.length; i++) {
boolean[] ifPrint = new boolean[nums.length];
printSubSet(nums, ifPrint, 0, i); // start from 0th index, and the size varies for the loop
}
}
// step 2: Recursively process elements from left to right to print out all possible combination of elements with fixed size
// we need three additional variables to keep track of status
// boolean array to know whether printed out or not
// start is the start index to be printed to prevent duplicates
// remain is keeping track of how many remaining elements to be processed for the subset action
public static void printSubSet(int[] nums, boolean[] ifPrint, int start, int remain) {
// firstly if remain == 0, we done!
if (remain == 0) {
System.out.print("{");
// check each ifPrint status to decide print or not
for (int i = 0; i < ifPrint.length; i++) {
if (ifPrint[i]) {
System.out.print(nums[i] + ", ");
}
}
System.out.print("}\n");
} else {
// we need process char by char from the start position until end
// before that, we need determine whether we proceed or not to check if (start + remain > nums.length)
if (start + remain > nums.length) {
} else {
for (int i = start; i < nums.length; i++) {
// now before we come to recursive part we have to make sure this position is not used
if (!ifPrint[i]) {
// now assign its value to true as used indicator
ifPrint[i] = true;
// recursive call
printSubSet(nums, ifPrint, i + 1, remain - 1);
// set the position back to false and proceed from next element
ifPrint[i] = false;
}
}
}
}
}
}
Console Output
