There is integer array like {1,2,4,5,6,1,2,4,3,5,7,2,1}. I want to find the possible combination of pair which sum is 4.
input : {1,2,4,5,6,1,2,4,3,5,7,2,1}
output : {1,1,2}, {2,2}, {3,1}, {1,2,1}...etc which make the sum as 4
A:
这是一个subset sum problem问题,该类型问题的解法见 wiki,比较复杂
http://en.wikipedia.org/wiki/Subset_sum_problem
code sample (passed testing) 该解法是暴力法,即递归求出所有可能的组合
View Code
import java.util.ArrayList;
import java.util.Arrays;
public class Test
{
public static void main(String[] args)
{
int[] theArray={1,2,4,5,6,1,2,4,3,5,7,2,1};
ArrayList<ArrayList<Integer>> allpairs=getPairs(theArray,0,4);
System.out.println(allpairs.size());
for(ArrayList<Integer> pair:allpairs)
{
System.out.println(Arrays.toString(pair.toArray()));
}
System.out.println("Complete the testing work!");
}
public static ArrayList<ArrayList<Integer>> getPairs(int[] theArray, int start, int sum)
{
if(sum==0)
{
ArrayList<ArrayList<Integer>> pairs=new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> pair=new ArrayList<Integer>();
pairs.add(pair);
return pairs;
}
if(start>=theArray.length)
{
return null;
}
ArrayList<ArrayList<Integer>> pairs=new ArrayList<ArrayList<Integer>>();
//situation 1 include the start element;
ArrayList<ArrayList<Integer>> subpairs=getPairs(theArray,start+1,sum-theArray[start]);
if(subpairs!=null)
{
for(ArrayList<Integer> subpair:subpairs)
{
ArrayList<Integer> pair=new ArrayList<Integer>();
if(pair!=null)
{
pair.add(theArray[start]);
pair.addAll(subpair);
pairs.add(pair);
}
}
}
//situation 2 do not include the start element;
ArrayList<ArrayList<Integer>> otherpairs=getPairs(theArray,start+1,sum);
if(otherpairs!=null)
{
for(ArrayList<Integer> otherpair:otherpairs)
{
pairs.add(otherpair);
}
}
return pairs;
}
}