LeetCode 039 Combination Sum

题目要求：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, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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]

代码如下：

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(candidates.begin(), candidates.end());
        vector<int>::iterator pos = unique(candidates.begin(), candidates.end());
        candidates.erase(pos, candidates.end());

        vector<vector<int> > ans;
        vector<int> record;
        searchAns(ans, record, candidates, target, 0);
        return ans;
    }

private:
    void searchAns(vector<vector<int> > &ans, vector<int> &record, vector<int> &candidates, int target, int idx) {

        if (target == 0) {
            ans.push_back(record);
            return;
        }

        if (idx == candidates.size() || candidates[idx] > target) {
            return;
        }

        for (int i = target / candidates[idx]; i >= 0; i--) {
            record.push_back(candidates[idx]);
        }

        for (int i = target / candidates[idx]; i >= 0; i--) {
            record.pop_back();
            searchAns(ans, record, candidates, target - i * candidates[idx], idx + 1);
        }
    }
};