Combination Sum

Given a set of candidate numbers (C) 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.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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] 

此题即给定数量的币值，如25分,10分,5分,1分，编写计算n分的有多少种表示方法变种

就是枚举，用DFS

每种情况可能：target/candidate[level] 去遍历 。注意的先排序保证非递减！

第一次：

#include <iostream>#include <vector>#include <algorithm>using namespace std;void combine_sum(vector<int> &candidate,int target,int level ,int sum,vector<int> res_one,vector<vector<int>> &res);vector<vector<int>>  combine_sum_final(vector<int> &candidate,int target);int main(){vector<int> candidate;int target;int temp;cin >> temp;while (temp >= 0){candidate.push_back(temp);cin >> temp;}cout<<endl;cin>>target; vector<vector<int>> res;res = combine_sum_final(candidate,target);cout<<"res:" <<res.size()<<endl;for (int i = 0 ;i < res.size();i++){for (int j = 0;j <res[i].size();j++){cout<< res[i][j]<<" ";}cout<<endl;}}vector<vector<int>>  combine_sum_final(vector<int> &candidate,int target){sort(candidate.begin(),candidate.end());if (target == 0 || candidate.size() == 0 || candidate[0] <= 0){return vector<vector<int>>();}vector<int>  res_one;vector<vector<int>> res;combine_sum(candidate,target,0,0,res_one,res); return res;}//需保证target 非0 ！！void combine_sum(vector<int> &candidate,int target,int level ,int sum,vector<int> res_one,vector<vector<int>> &res){if (level >= candidate.size()){if ( sum == target){res.push_back(res_one);}return ;}for (int i = 0;i <= ((target - sum)/candidate[level]); i++){sum = sum + candidate[level]*i;//同层会相互影响，在后面必须还原！！ if (sum > target){return ;}for (int j = 1;j <= i;j++){res_one.push_back(candidate[level]);//需在后面还原！} combine_sum(candidate,target,level+1,sum,res_one,res);for (int j = 1;j <= i;j++){res_one.pop_back();}sum = sum - candidate[level]*i;}}

第二次：

  vector<vector<int> > combinationSum(vector<int> &candidates, int target) {   sort(candidates.begin(),candidates.end()); if (target == 0 || candidates.size() == 0 || candidates[0] <= 0){return vector<vector<int>>();}vector<int>  res_one;vector<vector<int>> res;//combine_sum(candidates,target,0,0,res_one,res);combine_sum(candidates,target,0,res_one,res); return res;}void combine_sum(vector<int> &candidate,int target,int level ,vector<int> res_one,vector<vector<int>> &res){if (target < 0){return ;}if (level >= candidate.size()){if (target == 0){res.push_back(res_one);} return ;}for (int i = 0;i <= (target/candidate[level]); i++){for (int j = 1;j <= i;j++){res_one.push_back(candidate[level]);}combine_sum(candidate,target - candidate[level]*i,level+1,res_one,res);for (int j = 1;j <= i;j++){res_one.pop_back();} }}