题意

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.

{2:"abc", 3:"def", 4:"ghi", 5:"jkl", 6:"mno", 7:"pqrs", 8:"tuv", 9:"wxyz"}

For example:

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

输入一个由数字字符组成的字符串，求将这串数字作为输入时，“9键输入法”所能得到的所有情况。

思路

创建一个数组 vector v，v[i] 代表第 i 个数所可能产生的字母集。再根据 v 用类似建立多叉树的方法得到每一个解。

代码

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class Solution {
public:
    vector<string> bfs(vector<string>& tree)
    {
        vector<string> ret;
        queue<string> Q;
        for(int i = 0; i < (int)tree[0].size(); i++)
        {
            string str(1, tree[0][i]);
            Q.push(str);
        }
        while(!Q.empty())
        {
            string str = Q.front();
            Q.pop();
            int len = str.length();
            if(len < (int)tree.size())
            {
                for(int i = 0; i < (int)tree[len].size(); i++)
                {
                    string s(1, tree[len][i]);
                    str.append(s);
                    Q.push(str);
                    str.erase(str.end() - 1);
                }
            }
            else
            {
                ret.push_back(str);
            }
        }
        return ret;
    }
    vector<string> letterCombinations(string digits)
    {
        string dct[] = {"", "", "abc", "def", "ghi", "jkl", "nmo", "pqrs", "tuv", "wxyz"};
        vector<string> tree;
        vector<string> ret;
        for(int i = 0 ;i < (int)digits.size() ;i++)
        {
            tree.push_back(dct[digits[i] - '0']);
        }
        if(tree.size() > 0) ret = bfs(tree);
        return ret;
    }
};