代码随想录算法训练营day 25|第七章 回溯算法part02

216.组合总和III 

class Solution {
public:
    vector<vector<int>> res;
    vector<int> outcome;
    void backtracking(int k,int n,int begin){
        if(n==0&&k==0){
            res.push_back(outcome);
            return;
        }else if(k<=0||n<=0) return;
        for(int i=begin;i<10;i++){
            if(i>n) return;
            outcome.push_back(i);
            backtracking(k-1,n-i,i+1);
            outcome.pop_back();
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        backtracking(k,n,1);
        return res;
    }
};

这道题其实也算比较简单的，算是基础题目了，其实就是每次从对应字符串中选一个字母来构成整个字符串。注意是直接传入参数digits字符串，这样只要同时传入要处理的下标，就可以一点一点处理整个字符串了。

class Solution {
private:
    const string letterMap[10]{
        "",    //0
        "",    //1
        "abc", //2
        "def", //3
        "ghi", //4
        "jkl", //5
        "mno", //6
        "pqrs", //7
        "tuv", //8
        "wxyz", //9
    };
public:
    vector<string> res;
    string s;
    void backtracking(int index,string digits){
        if(index==digits.size()){
            res.push_back(s);
            return;
        }
        string letter=letterMap[digits[index]-'0'];
        for(int i=0;i<letter.size();i++){
            s.push_back(letter[i]);
            backtracking(index+1,digits);
            s.pop_back();
        }
    }
    vector<string> letterCombinations(string digits) {
        if(digits.size()==0) return res;
        backtracking(0,digits);
        return res;
    }
};