leetcode39 组合总和

参考：
https://leetcode.com/problems/combination-sum/discuss/16530/Concise-Backtracking-Solution

class Solution {
public:

    void search(vector<int>& num, int next, vector<int>& pSol, int target, vector<vector<int> >& result)
    {
        if(target == 0) //找到满足条件的路径
        {
            result.push_back(pSol);  //放入结果中
            return;
        }
        
        if(next == num.size() || target - num[next] < 0) //边界条件
            return;
            
        //对于每个元素，有两种处理方式，选当前元素或者不选当前元素
        pSol.push_back(num[next]); //选当前元素
        search(num, next, pSol, target - num[next], result); 
        pSol.pop_back(); //不选当前元素，将当前元素pop
        
        search(num, next + 1, pSol, target, result); //从下一个元素开始找
    }

    
    vector<vector<int> > combinationSum(vector<int> &num, int target) 
    {
        vector<vector<int> > result;
        sort(num.begin(), num.end());
        vector<int> pSol;
        search(num, 0, pSol, target, result);
        return result;    
    }
};

下面这个感觉写的不是很明白

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

采用深度搜索遍历，第一个for循环是该层的所有元素，对于每一个元素，先进行循环内的深度搜索，target<0算是一个剪枝条件。
转自
https://segmentfault.com/a/1190000007300946
参考
https://blog.youkuaiyun.com/jaster_wisdom/article/details/80683651
https://blog.youkuaiyun.com/qq_32805671/article/details/80783599
动态规划解法（未看）
https://leetcode.com/problems/combination-sum/discuss/16656/Dynamic-Programming-Solution

思路：
https://www.cnblogs.com/jimmycheng/p/7198156.html?utm_source=itdadao&utm_medium=referral