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);            

        }

    }

};