Combination Sum II —— LeetCode

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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  10,1,2,7,6,1,5  and target  8 , A solution set is:  [1, 7]   [1, 2, 5]   [2, 6]   [1, 1, 6]
前の問題と似ていますが、候補セットの要素は一度しか現れません.
問題解決の考え方:現在の要素の次からsumを計算し、candidateをリストに追加し、重複要素をスキップします.

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {

        List<List<Integer>> res = new ArrayList<>();

        if (candidates == null || candidates.length == 0) {

            return res;

        }

        Deque<Integer> tmp = new ArrayDeque<>();

        Arrays.sort(candidates);

        helper(res, tmp, 0, target, candidates);

        return res;

    }



    private void helper(List<List<Integer>> res, Deque<Integer> tmp, int start, int target, int[] candidate) {

        if (target == 0) {

            res.add(new ArrayList<>(tmp));

//            System.out.println(tmp);

            return;

        }

        for (int i = start; i < candidate.length && target >= candidate[i]; i++) {

            tmp.addLast(candidate[i]);

            helper(res, tmp, i + 1, target - candidate[i], candidate);

            tmp.removeLast();

            while (i < candidate.length - 1 && candidate[i + 1] == candidate[i]) {

                i++;

            }

        }

    }