这个可以用图的遍历做，用二维数组存图的邻接矩阵。将矩阵的row>col的元素填充1，其它为0，控制出现1+9 和 9+1重复的情况。

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

public class Test {
	public static int n = 10;
	public static int[] source = new int[n - 1];  //存储1到9的数字
	public static boolean[] visited = new boolean[n - 1];   //访问标记数组
	public static int metrix[][] = new int[n - 1][n - 1];   //邻接矩阵
	public static List<Integer> result = new ArrayList<Integer>();

	public static void main(String[] args) {
		for (int i = 0; i < n - 1; i++) {
			source[i] = i + 1; // 填充数字1....9
			for (int j = 0; j < n - 1; j++) {
				if (i > j)
					metrix[i][j] = 1; // 填充连接矩阵下半边
			}
		}
		for (int i = 0; i < n - 1; i++) {
			depthFirstSearch(i);      //从source[0]开始图的深度优先遍历
		}
	}

	public static void depthFirstSearch(int index) {
		visited[index] = true;
		result.add(source[index]);
		if (getSum() == n) {
			displaySolution();
		} else if (getSum() > n) {
			result.remove(result.size() - 1);
			visited[index] = false;
			return;
		} else if (getSum() < n) {
			for (int i = 0; i < n - 1; i++) {
				if (metrix[index][i] == 1 && visited[i] == false) {
					depthFirstSearch(i);  
				} else
					continue;
			}
		}
		result.remove(result.size() - 1);
		visited[index] = false;
	}

	public static int getSum() {
		int sum = 0;
		for (Integer o : result) {
			sum += o;
		}
		return sum;
	}

	public static void displaySolution() {
		for (Integer o : result)
			System.out.print(o + "  ");
		System.out.println();
	}
}

算法设计上不是有个例子么 完全一样

递归应该是可以解决的吧，下面的代码测试ok ，仅供参考～
public class JTest {
public static void dpxy(int x, int y, String result){
    if(result == "")
       result = x + " =";
    if(x < y || x <= 0)
      return;
    if(y == 1 || x == y){
         if(y == 1){
            for(int i=0;i<x;i++)
                 if(i == x-1)
                   result += "1";
                 else
                   result += "1+";
     }else{
          result += x;
     }
   System.out.println(result);
   return;
  }
  result += y + "+";
  for(int i=0;i<y;i++){
    dpxy(x - y, y-i, result);
  }
}
/**
  * @param args
  */
public static void main(String[] args) {
  for(int i=0;i<20;i++)   //n=20
     dpxy(20, 20-i, "");
}
}

较高效的计算算法

public class IntegerSumArrange {
	
	private int param;
	
	private int[] array;
	
	public void init(int param) {
		
		this.param = param;
	
		array = new int[param - 1];
		
		for (int i = 0; i < array.length; i++) {
			
			array[i] = i + 1;
		}
	}
	
	public void exhaustion() {
		
		Stack<Integer> stack = new Stack<Integer>();
		
		for (int i = 0; i < array.length; i++) {
			
			stack.push(array[i]);
			
			exhaustion(stack, array, i);
			
			stack.pop();
		}		
	}
	
	public void exhaustion(Stack<Integer> stack, int[] array, int index) {
		
		for (int i = index + 1; i < array.length; i++) {
			
			int value = calc(stack, array[i]);
			
			if (value == param) {
				
				print(stack, array[i]);
				
			} else if (value < param) {
				
				stack.push(array[i]);
				
				exhaustion(stack, array, i);
				
				stack.pop();
			}
		}
	}
	
	public void print(Stack<Integer> stack, int value) {
		
		for (Iterator<Integer> it = stack.iterator(); it.hasNext();) {
			
			System.out.print(it.next());
			
			System.out.print("+");
		}
		
		System.out.print(value);
		
		System.out.print("=");
		
		System.out.println(param);
	}
	
	public int calc(Stack<Integer> stack, int value) {
		
		for (Iterator<Integer> it = stack.iterator(); it.hasNext();) {
			
			value += it.next();
		}
		
		return value;
	}	

	public static void main(String[] args) throws Exception {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		System.out.print("please input an integer : ");
		
		Integer number = Integer.parseInt(br.readLine());
		
		IntegerSumArrange isa = new IntegerSumArrange();
		
		isa.init(number);
		
		isa.exhaustion();
	}
}

import java.util.*;

public class MyRecursion{
public void test(List list,int prefix,String strPrefix,int m){
if(prefix == 10)
System.out.println(strPrefix.trim());
for(int i=m;i<list.size()&& prefix <= 10;i++){
List<Integer> temp = new ArrayList<Integer>(list);
int a = temp.remove(i);
test(temp,prefix + a ,strPrefix+" "+a,i);
}
}

public static void main(String[] args){
MyRecursion mr = new MyRecursion();
List<Integer> list = new ArrayList<Integer>();
list.add(1);
list.add(2);
list.add(3);
list.add(4);
list.add(5);
list.add(6);
list.add(7);
list.add(8);
list.add(9);
list.add(10);
//list.add(11);
mr.test(list,0,"",0);
}
}code]

import java.util.Collections;
import java.util.LinkedList;

public class test {
	static LinkedList<Integer> list = new LinkedList<Integer>();

	static void getSum(int sum, int n) {
		if (n <= 0 || sum <= 0)
			return;
		if (sum < n) {
			getSum(sum, sum);
			return;
		}
		list.push(n);
		if (sum == n) {
			Collections.reverse(list);
			System.out.println(list.toString());
			Collections.reverse(list);
		}
		getSum(sum - n, n - 1);
		list.pop();
		getSum(sum, n - 1);
	}

递归的效率很低，1s钟处理100已经差不多到极限了，用动态规划吧。

收益匪浅呀

其实就是中兴的那道面试题，在中兴的这道题中，令n=m即为lz所求。

package a.test;

import java.util.Stack;

public class ZhongXing {

	/**
	 *  输入两个整数 n 和 m ，从数列 1 ， 2 ， 3.......n 中随意取几个数 , 使其和等于 m , 要求将其中所有的可能组合列出来
	 */
	public static void main(String[] args) {
		Stack<Integer> stack=new Stack<Integer>();
		find(10,10,stack);
	}

	public static void find(int sum,int n,Stack<Integer> stack){
		if(sum<=0||n<=0)return;
		if(sum==n){
			System.out.print(n+" ");
			printIntStackNoPop(stack);
			System.out.println();
		}
		stack.push(n);
		find(sum-n,n-1,stack);
		stack.pop();
		find(sum,n-1,stack);
	}
	
	public static void printIntStackNoPop(Stack<Integer> stack){
		for(int each:stack){
			System.out.print(each+ " ");
		}
	}
}