leetcode题解-39. Combination Sum

题目：

Given a set of candidate numbers (C) (without duplicates) 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.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]]

本题是寻找数组中和为target的不同组合，其中每个数字都可以用很多次。首先想到的是把数组先进行排序，然后使用backtracking的方法对其进行回溯法求结果，这里使用的回溯法和SubSet求子集的题目差不多。代码入下：

    public List<List<Integer>> combinationSum(int[] candidates, int target) {        List<List<Integer>> res = new ArrayList<>();        Arrays.sort(candidates);        dfs(candidates, res, new ArrayList<>(), target, 0);        return res;    }    public void dfs(int[] candidates, List<List<Integer>> res, List<Integer> tmp, int target, int idx){        if(target == 0)            //寻找到了目标，将结果保存到res链表中            res.add(new ArrayList<>(tmp));        else if(target < 0)            return;        else{            for(int i=idx; i<candidates.length; i++){                tmp.add(candidates[i]);                dfs(candidates, res, tmp, target-candidates[i], i);                //dfs执行完之后，也就是target<0直接返回，所以应该讲tmp最后一个元素删除                tmp.remove(tmp.size()-1);            }        }    }