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();

        }

    }