【DFS】LeetCode 17. Letter Combinations of a Phone Number-优快云博客

本文链接：https://blog.youkuaiyun.com/Allenlzcoder/article/details/80712396

本文详细介绍了LeetCode上的第17题Letter Combinations of a Phone Number的两种解决方案，包括回溯+递归的方法，并分析了其时间复杂度为O(3^n)，空间复杂度为O(n)。

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LeetCode 17. Letter Combinations of a Phone Number

Solution1：我的答案

利用8皇后同样的方法，回溯+递归
时间复杂度 $O(3^n)$ ，空间复杂度 $O(n)$
1.1写的略复杂

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0) return vector<string>();
        set<string> result; //自动避免重复
        unordered_map<char, vector<string> > num = {{'2', {"a", "b", "c"}},    
                                                      {'3', {"d", "e", "f"}}, 
                                                      {'4', {"g", "h", "i"}}, 
                                                      {'5', {"j", "k", "l"}}, 
                                                      {'6', {"m", "n", "o"}}, 
                                                      {'7', {"p", "q", "r", "s"}}, 
                                                      {'8', {"t", "u", "v"}}, 
                                                      {'9', {"w", "x", "y", "z"}}
                                                     };
        string temp = ""; 
        int start = 0;
        Combinate(num, result, temp, digits, start);
        return vector<string> (result.begin(), result.end());
    }

    void Combinate(unordered_map<char, vector<string>>& num, 
                   set<string>& result, string temp, string& digits, int start) {
        if (start == digits.size()) {
            result.insert(temp);
            return;
        }
        else {
            for (int i = 0; i < num[digits[start]].size(); i++) {
                temp = temp + num[digits[start]][i];
                Combinate(num, result, temp, digits, start + 1);
                temp.erase(temp.size() - 1);
            }
        }
    }
};

1.2代码略微简化一点

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.size() == 0) return vector<string>();
        set<string> result; //自动避免重复
        vector<string> num = {"abc", "def", "ghi", "jkl", 
                              "mno", "pqrs", "tuv", "wxyz"};
        string temp = ""; 
        int start = 0;
        Combinate(num, result, temp, digits, start);
        return vector<string> (result.begin(), result.end());
    }

    void Combinate(vector<string>& num, set<string>& result, 
                   string temp, string& digits, int start) {
        if (start == digits.size()) {
            result.insert(temp);
            return;
        } else {
            for (int i = 0; i < num[digits[start] - '2'].size(); i++) {
                temp = temp + num[digits[start] - '2'][i];
                Combinate(num, result, temp, digits, start + 1);
                temp.erase(temp.size() - 1);
            }
        }
    }
};

几个重要注意事项

1.C++的string对象利用下标操作符“[]”得到的数据类型是char，要初始化char类型的数据要用单引号修饰单个字符；
2.编译错误stray ‘\343’ in program解决办法
这种错误是由于代码中含有中文的引号or其他的全脚符号引起的，而小编遇到的是由中文空格引起的，不易发现
参考链接：https://blog.youkuaiyun.com/u012349696/article/details/51043830