https://leetcode.com/problems/combination-sum/

Given a set of candidate numbers (C) (without duplicates) 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. 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] ]

解题 ：subset的加强版，依然要backtrack所有组合,找出符合条件的组合.用target减去当前值来维护remain的值，当remain为0我们知道找到了一个valid combination。

public List<List<Integer>> combinationSum(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, new ArrayList<>(), nums, target, 0);
    return list;
}

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{ 
        for(int i = start; i < nums.length; i++){
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elements
            tempList.remove(tempList.size() - 1);
        }
    }
}