推荐先看我的一篇介绍Set去重的博文地址是 http://blog.csdn.net/bug_moving 看了这个之后，再来看下面的程序基本就能看懂了

题目

我也不太记得，因为是朋友给我口述的，然后给了我一个截图，看了图片大致也能知道题目要我们做什么

package yn; import java.util.ArrayList; import java.util.Collections; import java.util.HashSet; import java.util.Set; class Composition extends ArrayList<Integer> { @Override public boolean equals(Object other) { Composition comp = (Composition) other; Collections.sort(this); Collections.sort(comp); if (this.isEmpty() || comp.isEmpty() || this.size() != comp.size()) return false; for (int i = 0; i < this.size(); i++) if (this.get(i) != comp.get(i)) return false; return true; } @Override public int hashCode() { return 0; } } /** * 用递归法,比如2=1+1,3=1+(2的所有组成法),5需要分解1+4,2+3,因为3+2和2+3是一样的,for循环只要到i<=n/2就够了. 然后就是剔除1+1+2和1+2+1的情况,继承set的特性重写了Composition(每个拆分的方式)的equals. 懒得读取n值了,直接在main里面赋值给n * @author yxx * */ public class lhy { public static void main(String[] args) { int n = 5; System.out.println(toStr(calc(n))); } public static Set<Composition> calc(int n) { Set<Composition> possibility = new HashSet<Composition>(); Composition composition = new Composition(); switch (n) { case 1: composition.add(1); possibility.add(composition); return possibility; case 2: composition.add(1); composition.add(1); possibility.add(composition); return possibility; default: for (int i = 1; i <= n / 2; i++) { composition = new Composition(); composition.add(i); composition.add(n - i); possibility.add(composition); if (i <= n - i) { Set<Composition> partial_pos = calc(n - i); for (Composition pos : partial_pos) { pos.add(i); possibility.add(pos); } } } return possibility; } } public static String toStr(Set<Composition> possibility) { String str = "total : " + possibility.size() + "\n"; for (Composition pos : possibility) str += toStr(pos); return str; } public static String toStr(Composition composition) { String str = composition.get(0) + ""; for (int i = 1; i < composition.size(); i++) str += (" + " + composition.get(i)); str += "\n"; return str; } }

运行截图

至此这个问题基本解决，我也没有考虑效率问题啥，也不知道是不是效率超标了，先就这样吧。

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