Combination sum of leetcode topic analysis

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7,

A solution set is:

7

2, 2, 3

Using recursion , Note when saving results , We should build a new one ArrayList：result.add(new ArrayList< Integer>(cur)). otherwise result Will point to the ever-changing cur.

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        if (candidates == null || candidates.length == 0) {
            return new ArrayList<List<Integer>>();
        }
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        ArrayList<Integer> cur = new ArrayList<Integer>();
        Arrays.sort(candidates);
        dfs(0, target, result, cur, candidates);
        return result;
    }
    private void dfs(int start, int target, List<List<Integer>> result,
            ArrayList<Integer> cur, int[] candidates) {
        if (target == 0) {
            result.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, target - candidates[i], result, cur, candidates);
            cur.remove(cur.size() - 1);
        }
    }