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]

题目大意：跟上一题类似，但是有点区别就是候选集中的元素只能出现一次。

解题思路：每次从当前元素的下一个开始计算sum，并把candidate加入List，并且跳过重复元素。

    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++;

            }

        }

    }