这道题想到的就是dfs,在累加的和大于或等于target时到达递归树的终点。
代码如下:
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
int r=candidates.size()-1;
vector<vector<int>> res;
for(int i=r;i>=0;i--){
vector<int> tmp;
int sum=0;
dfs(candidates, tmp, res, target, i, sum);
}
return res;
}
void dfs(vector<int>& candidates, vector<int>& tmp, vector<vector<int>>& res, int target,int id,int& sum){
sum+=candidates[id];
tmp.push_back(candidates[id]);
if(sum>=target){
if(sum==target) res.push_back(tmp);
sum-=candidates[id];
tmp.pop_back();
return;
}
for(int i=id;i>=0;i--){
dfs(candidates,tmp,res,target,i,sum);
}
tmp.pop_back();
sum-=candidates[id];
}
};