17. Letter Combinations of a Phone Number

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

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给定一串手机按键数字序列，要求给出所有可能的字符串组合。本题实质是一个搜索问题，DFS算法如下：

class Solution {
public:
    void dfs(string& digits, int step, vector<string>& alphabeta, vector<string>& ans, bool is_first)
    {
        if (step == digits.size())
            return;
        string cur = alphabeta[digits[step] – ‘0’];
        if (is_first) {
            for (int i = 0; i < cur.size(); i++)
                ans.push_back(cur.substr(i, 1));
            is_first = false;
        }
        else {
            int sz = ans.size(); //size要提前抽出来
            for (int i = 0; i < sz; i++) {
                string tmp = ans[i];
                ans[i] = ans[i] + cur.substr(0, 1);
                for (int j = 1; j < cur.size(); j++)
                    ans.push_back(tmp + cur.substr(j, 1));
            }
        }
        dfs(digits, step + 1, alphabeta, ans, is_first);
    }
    vector<string> letterCombinations(string digits)
    {
        vector<string> alphabeta = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
        vector<string> ans;
        dfs(digits, 0, alphabeta, ans, true);
        return ans;
    }
};

本代码提交AC，用时0MS。
本题当然也可以用BFS做。

二刷：
原来解法有一个is_first，而且for循环里处理不够优雅，这一题明显可以用一种更漂亮、统一的方法，代码如下：

class Solution {
private:
    void dfs(string& digits, int step, vector<string>& alphabeta, vector<string>& ans, string& candidate)
    {
        if (step == digits.size()) {
            ans.push_back(candidate);
            return;
        }
        int idx = digits[step] – ‘0’;
        for (int i = 0; i < alphabeta[idx].size(); ++i) {
            candidate.push_back(alphabeta[idx][i]);
            dfs(digits, step + 1, alphabeta, ans, candidate);
            candidate.pop_back();
        }
    }
public:
    vector<string> letterCombinations(string digits)
    {
        if (digits == "")
            return {};
        vector<string> alphabeta = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
        vector<string> ans;
        string candidate = "";
        dfs(digits, 0, alphabeta, ans, candidate);
        return ans;
    }
};

本代码提交AC，用时0MS。