Combination Sum leetcode

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] 

這個題是一個NP問題，方法仍然是N-Queens中介紹的套路。基本思路是先排好序，然後每次遞歸中把剩下的元素一一加到結果集合中，並且把目標減去加入的元素，然後把剩下元素（包括當前加入的元素）放到下一層遞歸中解決子問題。算法複雜度因爲是NP問題，所以自然是指數量級的。注意在實現中for循環中第一步有一個判斷，那個是爲了去除重複元素產生重複結果的影響，因爲在這裏每個數可以重複使用，所以重複的元素也就沒有作用了，所以應該跳過那層遞歸。代碼如下： 

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > res;
        if(candidates.size() == 0) {
            return res;
        }
        sort(candidates.begin(),candidates.end());
        vector<int> tmp;
        findSum(res,candidates,tmp,0,target);
        return res;
    }
    void findSum(vector<vector<int> >& res,vector<int>& candidates,vector<int>& tmp,int start,int target) {
        if(target<0) {
            return;
        }else if(target == 0) {
            res.push_back(tmp);
            return;
        } else {
            for(int i=start;i<candidates.size();i++) {
                if(i>start && candidates[i] == candidates[i-1]) {
                    continue;
                }
                tmp.push_back(candidates[i]);
                findSum(res,candidates,tmp,i,target-candidates[i]);
                tmp.pop_back();
            }
        }
    }
};