Combination Sum III
题目描述:
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]]
题目大意:
给定数字k和n，从1～9中挑选k个数字，使得这k个数字的和等于n。求出所有符合条件的情况，同样相同元素只能使用一次。40. Combination Sum 的区别就是限制了k个数字和。
这道题和
题目代码:
class Solution {
public:vector<vector<int>> combinationSum3(int k, int n) {dfs(0,1,0,k,n);return ans;}
private:vector<vector<int>> ans;vector<int>elements;void dfs(int cnt, int cur, int sum, int k, int n){if(cnt == k && sum != n) return ;if(sum == n && cnt == k){ans.push_back(elements);return ;}for(int i = cur; i <= 9; i++){elements.push_back(i);dfs(cnt+1, i+1, sum+i, k, n); elements.pop_back();}}
};