216. Combination Sum III
Description
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.
Example 1:
Input: k = 3, n = 7
Output:
[[1,2,4]]
Example 2:
Input: k = 3, n = 9
Output:
[[1,2,6], [1,3,5], [2,3,4]]
Analysis
此题和之前的 Combination Sum一样，都是应用回溯法，不过需要加上一个sum参数记录当前的数组中的元素的和，其余没有任何变化。
代码如下:
code
class Solution {
private:
void backtracking(vector<vector<int>> & res,vector<int> & temp,vector<int> & base,int sum,int start,int k,int n){
if(sum==n&&temp.size()==k){
res.push_back(temp);
return;
}
for(int i=start;i<base.size();++i){
temp.push_back(base[i]);
backtracking(res,temp,base,sum+base[i],i+1,k,n);
temp.pop_back();
}
}
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<int> base={1,2,3,4,5,6,7,8,9};
vector<vector<int>> res;
vector<int> temp;
backtracking(res,temp,base,0,0,k,n);
return res;
}
};