Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations inC 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]

题目解析：

（1）主要是参考网上的相关实现。

（2）下面的实现方法利用了递归的思想，遍历相关的组合。

#include <iostream>#include <algorithm>#include <vector>using namespace std;void dfs(vector<int> &candidates,int size,int start,int sum,int target,vector<int> &save,vector<vector<int> > &res){if(sum == target){res.push_back(save);return;}for(int i=start;i<size;i++){if(sum + candidates[i] > target)return;elsesave.push_back(candidates[i]);dfs(candidates,size,i,sum+candidates[i],target,save,res);save.pop_back();}}vector<vector<int> > combinationSum(vector<int> &candidates, int target) {vector<vector<int> > res;vector<int> save;sort(candidates.begin(),candidates.end());dfs(candidates,candidates.size(),0,0,target,save,res);return res;}int main(void){vector<int> candidates;candidates.push_back(2);candidates.push_back(3);candidates.push_back(6);candidates.push_back(7);int target = 7;vector<vector<int> > res = combinationSum(candidates, target);for(int i=0;i<res.size();i++){vector<int> temp = res[i];for(int j=0;j<temp.size();j++)cout << temp[j] << " ";cout << endl;}system("pause");return 0;}