LeetCode(17)-Letter Combinations of a Phone Number

问题描述：

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"].

问题分析：

解法一：

枚举所有情况。

对于每一个输入数字，对于已有的排列中每一个字符串，分别加入该数字所代表的每一个字符。

所有是三重for循环。

举例：

初始化排列{""}

1、输入2，代表"abc"

已有排列中只有字符串""，所以得到{"a","b","c"}

2、输入3，代表"def"

(1)对于排列中的首元素"a"，删除"a"，并分别加入'd','e','f'，得到{"b","c","ad","ae","af"}

(2)对于排列中的首元素"b"，删除"b"，并分别加入'd','e','f'，得到{"c","ad","ae","af","bd","be","bf"}

(2)对于排列中的首元素"c"，删除"c"，并分别加入'd','e','f'，得到{"ad","ae","af","bd","be","bf","cd","ce","cf"}

注意

（1）每次添加新字母时，应该先取出现有ret当前的size()，而不是每次都在循环中调用ret.size()，因为ret.size()是不断增长的。

（2）删除vector首元素代码为：

ret.erase(ret.begin());

解法二：

深度优先搜索DFS

解题方法一：

class Solution {private:    const string dict[10] = {        " ",         "1", "abc", "def",        "ghi", "jkl", "mno",        "pqrs", "tuv", "wxyz",    };    public:    vector<string> letterCombinations(string digits) {        vector<string> ret;        if(digits == "") return ret;        ret.push_back("");                for( int i = 0; i < digits.size(); i++){            int size = ret.size();            for( int j = 0; j < size; j++){                auto cur = ret[0];                ret.erase( ret.begin() );                for( int k = 0; k < dict[digits[i] - '0'].size(); k++){                    ret.push_back( cur + dict[digits[i] - '0'][k]);                }            }        }        return ret;    }};

解法二：

class Solution {private:    const string alpha[10] = {        " ",        "1", "abc", "def",        "ghi", "jkl", "mno",        "pqrs", "tuv", "wxyz"    };    void dfs(vector<string> &res, string &ab, string &digits, int cur) {        if (cur >= digits.length()) {            res.push_back(ab);            return;        }        for (auto &a : alpha[digits[cur] - '0']) {            ab.push_back(a);            dfs(res, ab, digits, cur + 1);            ab.pop_back();        }    }public:    vector<string> letterCombinations(string digits) {        vector<string> res;        string alphas;        if(digits == "") return {};        dfs(res, alphas, digits, 0);        return res;    }};

解法二是看了github上的答案，但是对DFS具体用法还不熟悉，就要多看程序。