39. Combination Sum 凑出一个和，可以重复用元素（含duplicates）

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

可以不从第0位开始，直接从start开始。DFS中应该有start这个参数

循环里面当然是用i，不是start

需要考虑currentSum > target然后就会直接return ;的这种情况

和subset、全排列不同，这里的元素可以重复用。所以不需要

if(temp.contains(nums[i])) continue;

class Solution {
    public List<List<Integer>> combinationSum(int[] nums, int target) {
        //cc
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        if (nums == null || nums.length == 0)
            return results;
        
        //排序一下
        Arrays.sort(nums);
        
        backtrace(nums, new ArrayList<Integer>(), 0, 0,
                         target, results);
        
        return results;
    }
    
    public void backtrace(int[] nums, List<Integer> temp, int start, 
                          int currentSum,
                          int target, List<List<Integer>> results) {
        //exit
        if (currentSum == target) {
            results.add(new ArrayList<>(temp)); //必须这样写
        }else if (currentSum > target) {
            return ;
        }else {
            for (int i = start; i < nums.length; i++) {
                temp.add(nums[i]);
                backtrace(nums, temp, i, currentSum + nums[i],
                         target, results);
                temp.remove(temp.size() - 1);
            }
        }
    }
}