39. 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.
• 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]
]
采用深度优先搜索的思想，对于输入candidates=[1,2] ，target=3，遍历的方向如图： 

#include "stdafx.h"

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
void helper(int pos, int base, int target, vector<int>& candidates, vector<int>& path, vector<vector<int> >& result)
{
    if (base > target)
        return;
    if (base == target)
    {
        result.push_back(path);
        return;
    }
    for (int i = pos; i < candidates.size(); ++i)
    {
        path.push_back(candidates[i]);
        helper(i, base + candidates[i], target, candidates, path, result);
        path.pop_back();
    }

}
vector<vector<int> > combinationSum(vector<int>& candidates, int target)
{
    vector<vector<int> > result;
    vector<int> path;
    sort(candidates.begin(), candidates.end());
    vector<int>::iterator it = unique(candidates.begin(),candidates.end());
    candidates.erase(it, candidates.end());
    helper(0, 0, target, candidates, path, result);
    return result;
}
int main()
{
    vector<vector<int> > result;
    vector<int> candidates;
    candidates.push_back(2);
    candidates.push_back(2);
    candidates.push_back(3);
    candidates.push_back(7);
    int target = 7;
    result = combinationSum(candidates, target);

    return 0;
}