拆解数字

论坛上看到的一个题目：
http://www.iteye.com/topic/963980

将任一个数字进行拆解，例如：

3 = 2+1 = 1+1+1 所以3有三種拆法
4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 共五種
5 = 4 + 1 = 3 + 2 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 1 + 1 = 1 + 1 +1 +1 +1　共七种


随便给一个数字，对其进行拆解，并打印可拆解情况和拆解结果数。

public class SplitNum {
	public static void main(String[] args) {
		System.out.println("一共有" + splitNumber(10) + "种分法");
	}

	public static int splitNumber(int a) {
		return splitNumberWithMax(a, a, 0, a+" = ");
	}

	private static int splitNumberWithMax(int max, int tag, int c, String s) {
		if (tag == 1 || tag == 0) {
			System.out.println(tag == 1 ? (s + tag ) : s.substring(0,s.length()-2));
			c++;
		} else {
			for (int i = (max >= tag ? tag : max); i > 0; i--) {
				c = splitNumberWithMax(i, tag - i, c, s + i + " + ");
			}
		}
		return c;
	}
}

利用循环，不用递归，不用怎么担心内存的问题。

import java.util.Stack;

public class SplitNum2 {
	public static void main(String[] args) {
		System.out.println(splitNum(200));
	}

	public static int splitNum(int tag){
		int a,b;
		int c = 0;
		Stack<Integer> s = new Stack<Integer>();
		s.push(tag);
		while(true){
			c++;
			System.out.print(c+":");
			print(s);
			while (!s.isEmpty() && s.peek() == 1) {
				s.pop();
			}
			if (s.isEmpty())break;
			a = s.pop();
			s.push(--a);
			do {
				b = tag - sum(s);
				if (b==0)break;
				s.push(a<b?a:b);
			} while (true);
		}
		return c;
	}
	private static int sum(Stack<Integer> s){
		int sum = 0;
		for (int i:s) {
			sum += i;
		} 
		return sum;
	}

	private static void print(Stack<Integer> s){
		for (int i:s) {
			System.out.print(i+"+");
		} 
		System.out.println();
	}

}