17. Letter Combinations of a Phone Number LeetCode题解

Given a digit string, 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.

Input:Digit string "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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题意：

给定一个数字字符串，返回其数字串代表的所有可能字母组合；

数字和字符的映射关系如图所示；

题解：

2代表abd，3代表def，23则代表abc和def的所有组合，即如样例所示；

思路上可以使用递归或者循环，这里推荐写循环，因为面试官往往不喜欢递归；

Code【Java】

public class Solution {
    public List<String> letterCombinations(String digits) {
        if (digits.length() < 1) {
            return new LinkedList<String>();
        }
        String mapping[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        LinkedList<String> ans = new LinkedList<String>(Arrays.asList(""));
        for (int i = 0; i < digits.length(); ++i) {
            int num = digits.charAt(i) - '0';
            while (ans.peek().length() == i) {
                String cur = ans.poll();
                for (int j = 0; j < mapping[num].length(); ++j) {
                    ans.add(cur + mapping[num].charAt(j));
                }
            }
        }
        return ans;
    }
}

Code【C++】

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.size() < 1) {
            return vector<string>();
        }
        string mapping[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        queue<string> ans;
        ans.push("");
        for (int i = 0; i < digits.size(); ++i) {
            int num = digits[i] - '0';
            while (ans.front().size() == i) {
                string cur = ans.front();
                for (int j = 0; j < mapping[num].size(); ++j) {
                    ans.push(cur + mapping[num][j]);
                }
                ans.pop();
            }
        }
        vector<string> ret;
        while (!ans.empty()) {
            ret.push_back(ans.front());
            ans.pop();
        }
        return ret;
    }
};