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++;
            }
        }
    }