day25 - 回溯算法part02

216. 组合总和 III

详解

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    int sum = 0;

    void dfs(int k, int n, int start){
        if(path.size() == k){
            if(sum == n) result.push_back(path);
            return;
        }
        for(int i=start; i<=9; i++){
            path.push_back(i);
            sum += i;
            dfs(k, n, i+1);
            path.pop_back();
            sum -= i;
        }
    }

    vector<vector<int>> combinationSum3(int k, int n) {
        dfs(k, n, 1);
        return result;
    }
};

 

17. 电话号码的字母组合

class Solution {
public:
    string map_letter[10] = {
        "", // 0
        "", // 1
        "abc", // 2
        "def", // 3
        "ghi", // 4
        "jkl", // 5
        "mno", // 6
        "pqrs", // 7
        "tuv", // 8
        "wxyz", // 9
    };

    vector<string> result;
    string path;

    void dfs(string& digits, int idx){
        if(path.size() == digits.size()){
            result.push_back(path);
            return;
        }
        string letters = map_letter[digits[idx] - '0'];
        for(int i=0; i<letters.size(); i++){
            path.push_back(letters[i]);
            dfs(digits, idx+1);
            path.pop_back();
        }
    }

    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0) {
            return result;
        }
        dfs(digits, 0);
        return result;
    }
};