Combination Sum

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] 

Analysis: This problem is the same as Coin Change Problem (See Cracking chapter 9). In order to meet the requirement of this problem, we need to sort the candidates array in ascending order first and then solve it recursively. 

f(target) = f(target - 0*first_element) + f(target - 1*first_element) + f(target - 2*first_element) + ... f(target - n*first_element)

public class Solution {    public void combinationSum(int[] candidates, int index, int target, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> tem) {if(target == 0) {ArrayList<Integer> clone = new ArrayList<Integer>(tem);res.add(clone);return;}int i = 0;while(index<candidates.length && target >= i*candidates[index]) {for(int j=0; j<i; j++) {tem.add(candidates[index]);}combinationSum(candidates, index+1, target-i*candidates[index], res, tem);for(int k=0; k<i; k++) {tem.remove(tem.size()-1);}i++;}return;}public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();ArrayList<Integer> tem = new ArrayList<Integer>();Arrays.sort(candidates);combinationSum(candidates, 0, target, res, tem);return res;    }}