LeetCode OJ：Combination Sum III（组合之和III）

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

就是去k个数的和加起来等于n的组合的个数，数字只能取1-9，做法比较简单就是DFS吗，这里不再赘述，我比较喜欢用private变量，这样函数的参数式写出来比较简单：

 class Solution {
 public:
     vector<vector<int>> combinationSum3(int k, int n){
         size = k;
         target = n;
         vector<int> tmpVec;
         tmpVec.clear();
         ret.clear();
         dfs(tmpVec, target, );
         return ret;
     }
     void dfs(vector<int>& vec, int left, int start)
     {
         if(left < ) return;
         if(left ==  && vec.size() == size){
             ret.push_back(vec);
             return;
         }
         for(int i = start; i <= ; ++i){
             vec.push_back(i);
             dfs(vec, left - i, i + );
             vec.pop_back();
         }
     }
 private:
     vector<vector<int>> ret;
     int target;
     int size;
 };

java版本的如下所示，方法仍然相同，简单的DFS：

 public class Solution {
     public List<List<Integer>> combinationSum3(int k, int n) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         List<Integer> tmp = new ArrayList<Integer>();
         dfs(ret, tmp, 0, k, 1, n);
         return ret;
     }

     public void dfs(List<List<Integer>>ret, List<Integer>tmp, int dep, int maxDep, int start, int left){
         if(left < 0)
             return; //回溯
         else if(left == 0){
             if(dep == maxDep)
                 ret.add(new ArrayList<Integer>(tmp));
             return;
         }else{
             for(int i = start; i <= 9; ++i){
                 tmp.add(i);
                 dfs(ret, tmp, dep+1, maxDep, i+1, left - i);
                 tmp.remove(new Integer(i));
             }
         }
     }
 }