Combination sum II of leetcode topic analysis

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

1. Combination Sum Each element can be used more than once , So recursion is dfs(i, target - candidatesi, result, cur, candidates);
2. Combination Sum II Each element can only be used once , So recursion is dfs(i + 1, target - candidatesi, rt, cur, candidates); Because the solution may be repeated , So use set, Finally, it turns into list.

    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        if (candidates == null || candidates.length == 0) {
            return new ArrayList<List<Integer>>();
        }
        Set<List<Integer>> rt = new HashSet<List<Integer>>();
        ArrayList<Integer> cur = new ArrayList<Integer>();
        Arrays.sort(candidates);
        dfs(0, target, rt, cur, candidates);
        return new ArrayList<List<Integer>>(rt);
    }
    private void dfs(int start, int target, Set<List<Integer>> rt,
            ArrayList<Integer> cur, int[] candidates) {
        if (target == 0) {
            rt.add(new ArrayList<Integer>(cur));
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            // candidates[i] > target, Then the recursion ends , There can be no solution 
            if (candidates[i] > target) {
                return;
            }
            cur.add(candidates[i]);
            dfs(i + 1, target - candidates[i], rt, cur, candidates);
            cur.remove(cur.size() - 1);
        }
    }