LeetCode （17）Letter Combinations of a Phone Number

（17）Letter Combinations of a Phone Number

题目：通过所给字符串，返回按下数字可能返回的字符串组合，数字和字符的关系如同手机键盘一样。

例子：

输入：字符串为“23”。输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]。

首先想到的就是队列，不停地通过队列的每一组字符串相加获得下一组字符串，然后再把下一组字符串压入队列中，这样能够实现整个字符串的所有可能性的罗列。

当然在C++中vector有着类似队列和栈的功能，每一次拿出v.back()做基础字符串，然后v.pop_back()，直到v.empty()，将做好的新字符串放到v_temp中，最后v=v_temp，慢慢将所有的字符串处理完成就可以了。

下面是代码：

class Solution {public:    vector<string> letterCombinations(string digits) {        int len_str = digits.size();        char c;        string word[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};        int i,j;        vector<string> s;        int len = 0;        for(i = 0; i < len_str; i ++){            c = digits[i];            if(c == '1'){            }            else if(c == '0'){                vector<string> temp_v;                if(len == 0){                    temp_v.push_back(" ");                    s = temp_v;                    len ++;                }                else{                    while(!s.empty()){                        string str_temp = "";                        str_temp += s.back();                        s.pop_back();                        str_temp += " ";                        temp_v.push_back(str_temp);                    }                    s = temp_v;                }            }            else if(c == '2'||c == '3'||c == '4'||c == '5'||c == '6'||c == '8'){                vector<string> temp_v;                if(len == 0){                    string str_temp = "";                    string s1 = str_temp;                    s1.append(word[c - '0'].substr(0,1));                    temp_v.push_back(s1);                    string s2 = str_temp;                    s2.append(word[c - '0'].substr(1,1));                    temp_v.push_back(s2);                    string s3 = str_temp;                    s3.append(word[c - '0'].substr(2,1));                    temp_v.push_back(s3);                    s = temp_v;                    len ++;                }                else{                    while(!s.empty()){                        string str_temp = "";                        str_temp += s.back();                        s.pop_back();                        string s1 = str_temp;                        s1.append(word[c - '0'].substr(0,1));                        temp_v.push_back(s1);                        string s2 = str_temp;                        s2.append(word[c - '0'].substr(1,1));                        temp_v.push_back(s2);                        string s3 = str_temp;                        s3.append(word[c - '0'].substr(2,1));                        temp_v.push_back(s3);                    }                    s = temp_v;                }            }            else if(c == '7'||c == '9'){                vector<string> temp_v;                if(len == 0){                    string str_temp = "";                    string s1 = str_temp;                    s1.append(word[c - '0'].substr(0,1));                    temp_v.push_back(s1);                    string s2 = str_temp;                    s2.append(word[c - '0'].substr(1,1));                    temp_v.push_back(s2);                    string s3 = str_temp;                    s3.append(word[c - '0'].substr(2,1));                    temp_v.push_back(s3);                    string s4 = str_temp;                    s4.append(word[c - '0'].substr(3,1));                    temp_v.push_back(s4);                    s = temp_v;                    len ++;                }                else{                    while(!s.empty()){                        string str_temp = "";                        str_temp += s.back();                        s.pop_back();                        string s1 = str_temp;                        s1.append(word[c - '0'].substr(0,1));                        temp_v.push_back(s1);                        string s2 = str_temp;                        s2.append(word[c - '0'].substr(1,1));                        temp_v.push_back(s2);                        string s3 = str_temp;                        s3.append(word[c - '0'].substr(2,1));                        temp_v.push_back(s3);                        string s4 = str_temp;                        s4.append(word[c - '0'].substr(3,1));                        temp_v.push_back(s4);                    }                    s = temp_v;                }            }        }        return s;    }};