40. Combination Sum II

Backtracking

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Arrays.sort(candidates);
        backtracking(candidates, target, 0, 0);
        return result;
    }

    private void backtracking(int[] candidates, int target, int sum, int startIndex) {
        if (sum > target) return;
        if (sum == target) {
            result.add(new ArrayList<>(path));
            return;
        } 

        for (int i = startIndex; i < candidates.length; i++) {
            if (candidates[i] > target) break;
            if (i > startIndex && candidates[i] == candidates[i - 1]) continue;
            path.add(candidates[i]);
            backtracking(candidates, target, sum + candidates[i], i + 1);
            path.remove(path.size() - 1);
        }
    }
}

Remarks:

1. First sort array (very important!), then check the neighbor (next) candidate to avoid duplicates.
2. TC: $O(2^n*k)$ (k is the average length).
3. Same idea as LeetCode 39.