代码随想录算法训练营第25天｜216. 组合总和 III、17. 电话号码的字母组合-优快云博客

文章提供了两个编程问题的解决方案，一个是C++和Java版本的组合总和III问题，该问题涉及找到所有可能的整数组合，使得它们的和等于给定的目标值。另一个是电话号码的字母组合，将数字映射到字母并生成所有可能的字符串组合。这两个问题都使用了回溯算法来求解。

Day 25

216. 组合总和 III

class Solution {
public:
    vector<int> temp;
    vector<vector<int>> ans;

    bool check(int mask, int k, int n) {
        temp.clear();
        for (int i = 0; i < 9; ++i) {
            if ((1 << i) & mask) {
                temp.push_back(i + 1);
            }
        }
        return temp.size() == k && accumulate(temp.begin(), temp.end(), 0) == n; 
    }

    vector<vector<int>> combinationSum3(int k, int n) {
        for (int mask = 0; mask < (1 << 9); ++mask) {
            if (check(mask, k, n)) {
                ans.emplace_back(temp);
            }
        }
        return ans;
    }
};

java版：

class Solution {
    List<Integer> temp = new ArrayList<Integer>();
    List<List<Integer>> ans = new ArrayList<List<Integer>>();

    public List<List<Integer>> combinationSum3(int k, int n) {
        for (int mask = 0; mask < (1 << 9); ++mask) {
            if (check(mask, k, n)) {
                ans.add(new ArrayList<Integer>(temp));
            }
        }
        return ans;
    }

    public boolean check(int mask, int k, int n) {
        temp.clear();
        for (int i = 0; i < 9; ++i) {
            if (((1 << i) & mask) != 0) {
                temp.add(i + 1);
            }
        }
        if (temp.size() != k) {
            return false;
        }
        int sum = 0;
        for (int num : temp) {
            sum += num;
        }
        return sum == n;
    }
}

17. 电话号码的字母组合

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        vector<string> combinations;
        if (digits.empty()) {
            return combinations;
        }
        unordered_map<char, string> phoneMap{
            {'2', "abc"},
            {'3', "def"},
            {'4', "ghi"},
            {'5', "jkl"},
            {'6', "mno"},
            {'7', "pqrs"},
            {'8', "tuv"},
            {'9', "wxyz"}
        };
        string combination;
        backtrack(combinations, phoneMap, digits, 0, combination);
        return combinations;
    }

    void backtrack(vector<string>& combinations, const unordered_map<char, string>& phoneMap, const string& digits, int index, string& combination) {
        if (index == digits.length()) {
            combinations.push_back(combination);
        } else {
            char digit = digits[index];
            const string& letters = phoneMap.at(digit);
            for (const char& letter: letters) {
                combination.push_back(letter);
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.pop_back();
            }
        }
    }
};

Java版：

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> combinations = new ArrayList<String>();
        if (digits.length() == 0) {
            return combinations;
        }
        Map<Character, String> phoneMap = new HashMap<Character, String>() {{
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }};
        backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
        return combinations;
    }

    public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuffer combination) {
        if (index == digits.length()) {
            combinations.add(combination.toString());
        } else {
            char digit = digits.charAt(index);
            String letters = phoneMap.get(digit);
            int lettersCount = letters.length();
            for (int i = 0; i < lettersCount; i++) {
                combination.append(letters.charAt(i));
                backtrack(combinations, phoneMap, digits, index + 1, combination);
                combination.deleteCharAt(index);
            }
        }
    }
}