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] 

解题思路:用递归实现,题目要求：(a₁,a₂, … ,a_k) must be in non-descending order. (ie,a₁≤ a₂≤ … ≤ a_k).首先将列表排序.递归时下一个深度级的遍历开始位置由上一个深度级决定。用这样的方式保证要求.

 

#include<iostream>#include<vector>#include<algorithm>#include<numeric>using namespace std;void FindcombinationSum(vector<vector<int> >&ResultVector, vector<int> &candidates, vector<int> &Oneresult, int target, int num = 0){if (accumulate(Oneresult.begin(), Oneresult.end(), 0) >= target)return;for (int i = num; i != candidates.size();++i){Oneresult.push_back(candidates[i]);int sum = accumulate(Oneresult.begin(),Oneresult.end(),0);if (sum == target)ResultVector.push_back(Oneresult);FindcombinationSum(ResultVector, candidates, Oneresult, target, i);Oneresult.pop_back();}}vector<vector<int> > combinationSum(vector<int> &candidates, int target) {vector<vector<int> > ResultVector;vector<int>          Oneresult;sort(candidates.begin(), candidates.end());FindcombinationSum(ResultVector, candidates, Oneresult, target);return ResultVector;}