[LeetCode] Combination Sum III 组合之和之三
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.
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]]
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
这道题题是组合之和系列的第三道题,跟之前两道 Combination Sum,Combination Sum II 都不太一样,那两道的联系比较紧密,变化不大,而这道跟它们最显著的不同就是这道题的个数是固定的,为k。个人认为这道题跟那道 Combinations 更相似一些,但是那道题只是排序,对k个数字之和又没有要求。所以实际上这道题是它们的综合体,两者杂糅到一起就是这道题的解法了,n是k个数字之和,如果n小于0,则直接返回,如果n正好等于0,而且此时out中数字的个数正好为k,说明此时是一个正确解,将其存入结果res中,具体实现参见代码入下:
class Solution {
public:
vector<vector<int> > combinationSum3(int k, int n) {
vector<vector<int> > res;
vector<int> out;
combinationSum3DFS(k, n, , out, res);
return res;
}
void combinationSum3DFS(int k, int n, int level, vector<int> &out, vector<vector<int> > &res) {
if (n < ) return;
if (n == && out.size() == k) res.push_back(out);
for (int i = level; i <= ; ++i) {
out.push_back(i);
combinationSum3DFS(k, n - i, i + , out, res);
out.pop_back();
}
}
};
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