[LeetCode] 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] 

一开始想到用背包DP做，用DP的好处是速度快。但打印结果比较麻烦。DP比较适合打印总共有几种情况。所以就用DFS来做。

class Solution {
private:
    vector<vector<int> > ret;
    vector<int> a;
public:
    void solve(int dep, int maxDep, int target, vector<int> &cand)
    {
        if (target < 0)
            return;
            
        if (dep == maxDep)
        {
            if (target == 0)
            {
                vector<int> res;
                for(int i = 0; i < maxDep; i++)
                    for(int j = 0; j < a[i]; j++)
                        res.push_back(cand[i]);
                ret.push_back(res);
            }
            return;
        }
        
        for(int i = 0; i <= target / cand[dep]; i++)
        {
            a[dep] = i;
            solve(dep + 1, maxDep, target - cand[dep] * i, cand);
        }
    }
    
    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());
        
        a.resize(candidates.size());
        ret.clear();
        if (candidates.size() == 0)
            return ret;
            
        solve(0, candidates.size(), target, candidates);
        
        return ret;
    }
};