【算法训练营day25】LeetCode216. 组合总和III LeetCode17. 电话号码的字母组合

LeetCode216. 组合总和III

题目链接：216. 组合总和III

独上高楼，望尽天涯

继续复健，一直在犯小的语法错误。

慕然回首，灯火阑珊

和昨天的题很像，主要区别在于递归返回条件和回溯过程。

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(int k, int target_sum, int start_index) {
        if (path.size() == k) {
            if (target_sum == 0) result.push_back(path);
            return;
        }
        for (int i = start_index; i <= 9; i++) { // 注意中间是<=
            target_sum -= i;
            path.push_back(i);
            backtracking(k, target_sum, i+1);
            target_sum += i;
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);
        return result;
    }
};

剪枝操作

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(int k, int target_sum, int start_index) {
        if (target_sum < 0) { // 剪枝操作
            return;
        }
        if (path.size() == k) {
            if (target_sum == 0) result.push_back(path);
            return;
        }
        for (int i = start_index; i <= 9 - (k - path.size()) + 1; i++) { // 剪枝操作
            target_sum -= i;
            path.push_back(i);
            backtracking(k, target_sum, i+1);
            target_sum += i;
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);
        return result;
    }
};

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LeetCode17. 电话号码的字母组合

题目链接：17. 电话号码的字母组合

独上高楼，望尽天涯

思路不难，细节很多，写起来有点浆糊。

慕然回首，灯火阑珊

class Solution {
private:
    const string letter_map[10] = {
        "",
        "",
        "abc",
        "def",
        "ghi",
        "jkl",
        "mno",
        "pqrs",
        "tuv",
        "wxyz"
    };
    vector<string> result;
    string path;
    void backtracking(const string& digits, int index) {
        if (path.size() == digits.size()) {
            result.push_back(path);
            return;
        }
        int digit = digits[index] - '0'; // char转int小技巧
        string letter = letter_map[digit];
        for (int i = 0; i < letter.size(); i++) {
            path.push_back(letter[i]); // 处理
            backtracking(digits, index+1); //递归
            path.pop_back(); // 回溯
        }
    }
public:
    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0) {
            return result;
        }
        backtracking(digits, 0);
        return result;
    }
};