39. Combination Sum-dfs

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]]

同上一题的套路很像，但是不同的是允许重复

public class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        ArrayList<List<Integer>> finalans = new ArrayList<List<Integer>>();        ArrayList<Integer> curans = new ArrayList<Integer>();        int index=0;        if(candidates==null||candidates.length==0) return finalans;        findAns(candidates,target,index,finalans,curans);        return finalans;    }        private void findAns(int[] candidates,int target,int index,ArrayList<List<Integer>> finalans,ArrayList<Integer> curans){        if(target==0){            finalans.add(new ArrayList<Integer>(curans));            return;            //return后面也可以不带参数，不带参数就是返回空，其实主要目的就是用于想中断函数执行，返回调用函数处            //终止当前循环，也不执行remove了            //返回上一层循环        }        if(target<0) return;                for(int i=index;i<candidates.length;i++){            if(i>0 && candidates[i]==candidates[i-1]) continue;//如果遇到重复的元素 跳过不执行            curans.add(candidates[i]);            int sum=target-candidates[i];            findAns(candidates,sum,i,finalans,curans);//这里依旧是传入i而不是i+1因为允许一个元素重复使用            curans.remove(curans.size()-1);        }    }}