40. 组合总和 II

给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用一次。

说明：

所有数字（包括目标数）都是正整数。
解集不能包含重复的组合。 
示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
示例 2:

输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:
[
  [1,2,2],
  [5]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/combination-sum-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        
        size = len(candidates)
        if size == 0:
            return []
        
        candidates.sort()
        
        path = []
        res = []
        self._find_path(candidates, path, res, target, 0, size)
        
        return res
    
    def _find_path(self, candidates, path, res, target, begin, size):
        if target == 0:
            res.append(path.copy())
        else:
            for i in range(begin, size):
                left_num = target - candidates[i]
                if left_num < 0:
                    break

                if i > begin and candidates[i] == candidates[i-1]:
                    continue
                path.append(candidates[i])

                self._find_path(candidates, path, res, left_num, i+1, size)#这里需要i+1
                path.pop()

class Solution {
    public List<List<Integer>> combinationSum2(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++){
                if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
                tempList.add(nums[i]);
                backtrack(list, tempList, nums, remain - nums[i], i + 1);
                tempList.remove(tempList.size() - 1); 
            }
        }
    }
}

reference 

https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning)