求和数组项值，使其等于某个目标值

描述

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may
be chosen from C unlimited number of times.
Note: All numbers (including target) will be positive integers. Elements in a combination (a1, a2, ... , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). The
solution set must not contain duplicate combinations. For example, given candidate
set 2,3,6,7 and target 7, A solution set is:
[7]
[2, 2, 3]

代码

package com.lilei.spring_boot_db.boot.pack1115;

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

public class combination_sum {

	public static void main(String[] args) {
		int[] array = new int[]{2,3,6,7};
		com_sum(array,0,0,8,new ArrayList<Integer>());
	}
	
	static void com_sum(int[] array,int p,int value,int target,List<Integer> list){
		
		if (value + array[p] < target){
			
			List<Integer> dest1 = new ArrayList<Integer>();
			
			copyList(dest1,list);
			
			dest1.add(array[p]);
			
			com_sum(array,p,value+ array[p],target,dest1);
			if (p+1 < array.length) {
				List<Integer> dest2 = new ArrayList<Integer>();
				copyList(dest2,list);
				com_sum(array, p + 1, value, target,dest2);
			}
		}else if (value + array[p] == target){
			list.add(array[p]);
			
			for(int v:list)
				System.out.print(v+",");
			System.out.println();
			
		}
	}
	
	static void copyList(List<Integer> dest,List<Integer> src){
		for(int v:src)
			dest.add(v);
	}

}