17. Letter Combinations of a Phone Number

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Created At : 2021-04-07 23:39

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

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

Example 1:

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

Example 2:

**Input:** digits = ""
**Output:** []

Example 3:

**Input:** digits = "2"
**Output:** ["a","b","c"]

Constraints:

0 <= digits.length <= 4
digits[i] is a digit in the range ['2', '9'].

这道题让我们求电话号码的字母组合，即数字2到9中每个数字可以代表若干个字母，然后给一串数字，求出所有可能的组合，相类似的题目有 Path Sum II，Subsets II，Permutations，Permutations II，Combinations，Combination Sum 和 Combination Sum II 等等。这里可以用递归 Recursion 来解，需要建立一个字典，用来保存每个数字所代表的字符串，然后还需要一个变量 pos，记录当前生成的字符串的字符个数，实现套路和上述那些题十分类似。在递归函数中首先判断 pos，如果跟 digits 中数字的个数相等了，将当前的组合加入结果 res 中，然后返回。我们通过 digits 中的数字到 dict 中取出字符串，然后遍历这个取出的字符串，将每个字符都加到当前的组合后面，并调用递归函数即可，参见代码如下：

解法一：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res;
        vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        dfs(digits, dict, 0, "", res);
        return res;
    }
    void dfs(string digits, vector<string>& dict, int pos, string cur, vector<string>& res) {
        if (pos == digits.size()) { res.push_back(cur); return; }
        string str = dict[digits[pos] - '0'];
        for (int i = 0; i < str.size(); ++i) {
            dfs(digits, dict, pos + 1, cur + str[i], res);
        }
    }
};

这道题也可以用迭代 Iterative 来解，在遍历 digits 中所有的数字时，先建立一个临时的字符串数组t，然后跟上面解法的操作一样，通过数字到 dict 中取出字符串 str，然后遍历取出字符串中的所有字符，再遍历当前结果 res 中的每一个字符串，将字符加到后面，并加入到临时字符串数组t中。取出的字符串 str 遍历完成后，将临时字符串数组赋值给结果 res，具体实现参见代码如下：

解法二：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {};
        vector<string> res{""};
        vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        for (int i = 0; i < digits.size(); ++i) {
            vector<string> t;
            string str = dict[digits[i] - '0'];
            for (int j = 0; j < str.size(); ++j) {
                for (string s : res) t.push_back(s + str[j]);
            }
            res = t;
        }
        return res;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/17

类似题目：

Generate Parentheses

Combination Sum

Binary Watch

参考资料：

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

https://leetcode.com/problems/letter-combinations-of-a-phone-number/discuss/8109/My-recursive-solution-using-Java

https://leetcode.com/problems/letter-combinations-of-a-phone-number/discuss/8097/My-iterative-sollution-very-simple-under-15-lines

https://leetcode.com/problems/letter-combinations-of-a-phone-number/discuss/8207/Concise-15-line-Java-Iterative-Solution-very-Straight-Forward-with-Brief-Explanation

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