Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
All numbers (including target) will be positive integers.The solution set must not contain duplicate combinations.
For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is:
[ [7], [2, 2, 3] ] 本题宜采用回溯法，代码如下： #include<iostream> #include<vector> #include<algorithm> using namespace std; void backTracking(vector<vector<int>> &res, vector<int> arr ,vector<int>candidate ,int start,int target ) { if (target == 0)//满足条件，输出结果 { res.push_back(arr); return; } for (int i = start; i < candidate.size(); i++)//枚举所有可能的路径 { if (candidate[i] <= target)//满足界限函数和约束条件 { arr.push_back(candidate[i]); backTracking(res, arr, candidate, i, target - candidate[i]); arr.pop_back();//回溯清理工作 } } } int main() { vector<int> candidate = { 2, 3, 6, 7 }; int target = 7; vector<vector<int>> res; vector<int> arr; sort(candidate.begin(), candidate.end()); backTracking(res, arr, candidate, 0, target); for (int i = 0; i < res.size(); i++) { for (int j = 0; j < res[i].size(); j++) { cout << res[i][j] << " "; } cout << endl; } return 0; }
