216. 组合总和 III

方法：递归

class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void solve(int k, int goal, int cur, int idx) {
        if (cur > goal) return;
        if (path.size() == k) {
            if (cur == goal) res.push_back(path);
            return ;
        }

        for (int i = idx; i <= 9-(k-path.size()) + 1; ++i) {
            path.push_back(i);
            solve(k, goal, cur + i, i + 1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        res.clear();
        path.clear();
        solve(k, n, 0, 1);
        return res;
    }
};

$时间复杂度O(*k),空间复杂度(n)

17. 电话号码的字母组合

方法：递归

class Solution {
private:
    const string letterMap[10] = {
        "",
        "",
        "abc",
        "def",
        "ghi",
        "jkl",
        "mno",
        "pqrs",
        "tuv",
        "wxyz",
    };
    vector<string> res;
    string str;
    void solve(const string& digits, int idx, const string& str) {
        if (idx == digits.size()) {
            res.push_back(str);
            return ;
        }
        int num = digits[idx] - '0';
        string word = letterMap[num];
        for (int i = 0; i < word.size(); ++i) {
            solve(digits, idx + 1, str+word[i]);
        }
    }
public:
    vector<string> letterCombinations(string digits) {
        int n = digits.size();
        if (!n) return res;
        solve(digits, 0, str);
        return res;
    }
};

$时间复杂度O(),空间复杂度(n)