JAVA求集合中的组合

好几个月没弄代码了，今天弄个求组合的DEMO

思路是将集合的每个值对照一个索引，索引大小是集合的大小+2.索引默认为[000...000]，当组合后选取的组合值demo为[0100..00]。然后根据遍历索引来到集合中取值。

上代码：

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;

public class ComBit {

    public static void main(String[] args) {
        // Integer test
        int[] combination = new int[] { 1, 2, 3, 4 };
        List<Integer> combinationlist = new ArrayList<Integer>();
        for (Integer integer : combination) {
            combinationlist.add(integer);
        }
        ComBitIterator<Integer> i = new ComBitIterator<Integer>(combinationlist);
        while (i.hasNext()) {
            Object obj = i.next();
            System.out.println(obj.toString());
        }
        // String test
        String[] str = new String[] { "apple", "orange", "tomato", "potato" };
        List<String> combinationSlist = new ArrayList<String>();
        for (String s : str) {
            combinationSlist.add(s);
        }
        ComBitIterator<String> ii = new ComBitIterator<String>(combinationSlist);
        while (ii.hasNext()) {
            Object obj = ii.next();
            System.out.println(obj.toString());
        }

    }

}

class ComBitIterator<T> implements Iterator<T> {

    private int[] _bitArray = null;

    protected final int _length;

    protected final List<T> combination;

    protected List<T> _currentSet;

    public ComBitIterator(List<T> combination) {
        _currentSet = new ArrayList<T>();
        this._length = combination.size();
        this._bitArray = new int[_length + 2];
        this.combination = combination;
    }

    @Override
    public boolean hasNext() {
        return _bitArray[_length + 1] != 1;
    }

    @SuppressWarnings("unchecked")
    @Override
    public T next() {
        _currentSet.clear();
        for (int index = 1; index <= _length; index++) {
            if (_bitArray[index] == 1) {
                T value = combination.get(index - 1);
                _currentSet.add(value);
            }
        }
        int i = 1;
        while (_bitArray[i] == 1) {
            _bitArray[i] = 0;
            i++;
        }
        _bitArray[i] = 1;
        return (T) _currentSet;
    }

}

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PS:代码源于要求求数字1-20中 任取N的组合后合值为36，于是思路想到两种，第一种是递归，第二种虽然也是递归，但是想弄个通用的，就有了以上的代码。另外群里有人写出了直接递归的代码，也附上：

public static void main(String[] args) {
        combinateDataOfRange(1, 20, 36);
    }

    public static void combinateDataOfRange(int min, int max, int target) {
        combinateData(0, min, max, target, new Stack<Integer>());
    }

    public static void combinateData(int sum, int min, int max, int target,
            Stack<Integer> stack) {
        for (int i = stack.isEmpty() ? min : (Integer) stack.lastElement() + 1; i < max; ++i) {
            int tempSum = sum + i;
            stack.push(i);
            if (tempSum == target) {
                System.out.println(stack + "=" + target);
            } else {
                combinateData(tempSum, min, max, target, stack);
            }
            stack.pop();
        }
    }