题目描述

2.

以下是回溯算法的模版

class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;//这个变量名还是设为path更合适
    void backtrace(int n, int k, int startindex){
        //递归结束条件，这个是根据问题要求修改的
        if(path.size() == k){
            //找到了一个合适的组合
            res.push_back(path);
            return ;
        }

        //回溯过程
        for(int i = startindex; i <= n - (k - path.size()) + 1; i++){
            path.push_back(i);
            //向下递归的操作可以看作一个子问题
            backtrace(n,k,i+1);
            path.pop_back();
        }
    }
public:
        vector<vector<int>> combine(int n, int k) {
        res.clear();
        path.clear();
        backtrace(n,k,1);
        return res;
    }
};

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backtrace(int n, int k, int startindex){
        if(path.size() == k && accumulate(path.begin(), path.end(), 0) == n){
            res.push_back(path);
            return ;
        }

        for(int i = startindex; i <= 9; i++){
            path.push_back(i);
            backtrace(n,k,i+1);
            path.pop_back();
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        res.clear();
        path.clear();
        backtrace(n, k, 1);
        return res;
    }
};