我正在与我创建的,其中包含了一组数字作为数据的链接列表。我需要找到一个方法来测试这个名单的东西一切可能的两集的分区,而要做到这一点,我需要打破名单为每一个可能的两集组合。顺序并不重要,会出现重复。
例如,对于数字{1 4 3 1}的名单,可能的分裂是
{1}和{4,3,1}
{4}和{1,3,1}
{3}和{1,4,1}
{1}和{1,4,3}
{1,4}和{3,1}
{1,3}和{4,1}
{1,1}和{4,3}
4号的名单并不难,但事情变得更加复杂,因为列表变大,而我无法看到的模式。谁能帮我找到一个算法呢?
编辑:
对不起,我没有看到这个问题。这是我试过至今。我的循环结构是错误的。当我找出我想定期阵列后我在做,我将扩展算法,以适应我的链接列表。
公共类TwoSubsets
{
公共静态无效的主要(字串[] args)
{
INT []列表= {1,3,5,7,8};
INT地方= 1;
INT [] subsetA =新INT [10];
INT [] subsetB =新INT [10];
的for(int i = 0; I< list.length;我++)
{
subsetA [i] =列表[我]
为(中间体电流= 0;电流I;(5 - ⅰ);电流++)
{
subsetB [当前] =列表[地方]
地方++;
}
System.out.print(subsetA =);
对于(INT J = 0; J< subsetA.length; J ++)
{
System.out.print(subsetA [J]。+);
}
的System.out.println();
System.out.print(subsetB =);
对于(INT K = 0; K< subsetB.length; k ++)
{
System.out.print(subsetB [K] +);
}
}
}
}
解决方案
code:
公共静态无效的主要(字串[] args){
对于(字符串元素:findSplits(名单)){
的System.out.println(元);
}
}
静态的ArrayList<字符串> findSplits(ArrayList的<整数GT;集){
ArrayList的<字符串>输出=新的ArrayList();
ArrayList的<整数GT;第一=新的ArrayList(),第二=新的ArrayList();
串位串;
INT位=(int)的Math.pow(2,set.size());
而(bits--大于0){
比特串的String.Format =(%+ set.size()+的,Integer.toBinaryString(比特))代替('','0')。
的for(int i = 0; I< set.size();我++){
如果(bitString.substring(I,I + 1)的.equals(0)){
first.add(set.get(ⅰ));
} 其他 {
second.add(set.get(ⅰ));
}
}
如果(first.size()&其中; set.size()&安培;&安培; second.size()&其中; set.size()){
如果(output.contains(第一+!+二)及和放大器;!output.contains(第二++先)){
output.add(第一++二);
}
}
first.clear();
second.clear();
}
返回输出;
}
输出:
[1] [1,4,3]
[3] [1,4,1]
[3,1] [1,4]
[4] [1,3,1]
[4,1] [1,3]
[4,3] [1,1]
[4,3,1] [1]
这是沿着你要找的线路?如果没有,让我知道,我会做出调整或添加评论是必要的。
I am working with a linked list that I created which contains a set of numbers as data. I need to find a way to test every possible two-set partition of this list for something, and to do that, I need to break the list into every possible two-set combination. Order is not important and there will be duplicates.
For instance, for a list of numbers {1 4 3 1}, the possible splits are
{1} and {4, 3, 1}
{4} and {1, 3, 1}
{3} and {1, 4, 1}
{1} and {1, 4, 3}
{1, 4} and {3, 1}
{1, 3} and {4, 1}
{1, 1} and {4, 3}
A list of 4 numbers is not difficult, but things become more complicated as the list grows larger, and I am having trouble seeing a pattern. Can anyone help me find an algorithm for this?
Edit:
Sorry, I didn't see the question. This is what I have tried so far. My loop structure is wrong. When I figure out what I am doing after trying on a regular array, I will extend the algorithm to fit my linked list.
public class TwoSubsets
{
public static void main(String[] args)
{
int[] list = {1, 3, 5, 7, 8};
int places = 1;
int[] subsetA = new int[10];
int[] subsetB = new int[10];
for (int i = 0; i < list.length; i++)
{
subsetA[i] = list[i];
for (int current = 0; current < (5 - i ); current++)
{
subsetB[current] = list[places];
places++;
}
System.out.print("subsetA = ");
for (int j = 0; j < subsetA.length; j++)
{
System.out.print(subsetA[j] + " ");
}
System.out.println();
System.out.print("subsetB = ");
for (int k = 0; k < subsetB.length; k++)
{
System.out.print(subsetB[k] + " ");
}
}
}
}
解决方案
Code:
public static void main(String[] args) {
for(String element : findSplits(list)) {
System.out.println(element);
}
}
static ArrayList<String> findSplits(ArrayList<Integer> set) {
ArrayList<String> output = new ArrayList();
ArrayList<Integer> first = new ArrayList(), second = new ArrayList();
String bitString;
int bits = (int) Math.pow(2, set.size());
while (bits-- > 0) {
bitString = String.format("%" + set.size() + "s", Integer.toBinaryString(bits)).replace(' ', '0');
for (int i = 0; i < set.size(); i++) {
if (bitString.substring(i, i+1).equals("0")) {
first.add(set.get(i));
} else {
second.add(set.get(i));
}
}
if (first.size() < set.size() && second.size() < set.size()) {
if (!output.contains(first + " " + second) && !output.contains(second + " " + first)) {
output.add(first + " " + second);
}
}
first.clear();
second.clear();
}
return output;
}
Output:
[1] [1, 4, 3]
[3] [1, 4, 1]
[3, 1] [1, 4]
[4] [1, 3, 1]
[4, 1] [1, 3]
[4, 3] [1, 1]
[4, 3, 1] [1]
Is this along the lines of what you were looking for? If not, let me know and I will make adjustments or add comments as necessary.
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