Backtracking: 17. Letter Combinations of a Phone Number

写法一：

这种方法只能击败1%，但是经过小小的优化，性能有了很大的提高。

class Solution {public:    vector<string> rel;    string tmp;    int d_len;    char d[10][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    void findCombination(string a, int stt, string in_data) {        if(stt == d_len) {            rel.push_back(tmp);            return;        }        for(int i = 0; i < strlen(d[in_data[stt] - '0']); i++) {            string tmp2 = tmp;            tmp += d[in_data[stt] - '0'][i];            findCombination(tmp, stt + 1, in_data);            tmp = tmp2; // d[in_data[i] - '0'][j];        }        }    vector<string> letterCombinations(string digits) {        d_len = digits.length();        if(!d_len)            return rel;        findCombination(tmp, 0, digits);        return rel;    }};

写法二：

函数调用计算字符串长度写到循环外。这种写法击败了40%。

class Solution {public:    vector<string> rel;    string tmp;    int d_len;    char d[10][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    void findCombination(string a, int stt, string in_data) {        if(stt == d_len) {            rel.push_back(tmp);            return;        }        int di_len = strlen(d[in_data[stt] - '0']);        for(int i = 0; i < di_len; i++) {            string tmp2 = tmp;            tmp += d[in_data[stt] - '0'][i];            findCombination(tmp, stt + 1, in_data);            tmp = tmp2; // d[in_data[i] - '0'][j];        }        }    vector<string> letterCombinations(string digits) {        d_len = digits.length();        if(!d_len)            return rel;        findCombination(tmp, 0, digits);        return rel;    }};

写法三：

使用string的push_back和pop_back来操作，发现和速度特别慢。

class Solution {public:    vector<string> rel;    string tmp;    int d_len;    char d[10][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    void findCombination(string a, int stt, string in_data) {        if(stt == d_len) {            rel.push_back(tmp);            return;        }        int di_len = strlen(d[in_data[stt] - '0']);        for(int i = 0; i < di_len; i++) {            // string tmp2 = tmp;            tmp.push_back(d[in_data[stt] - '0'][i]); //  += d[in_data[stt] - '0'][i];            findCombination(tmp, stt + 1, in_data);            tmp.pop_back(); //tmp = tmp2; // d[in_data[i] - '0'][j];        }        }    vector<string> letterCombinations(string digits) {        d_len = digits.length();        if(!d_len)            return rel;        findCombination(tmp, 0, digits);        return rel;    }};

写法四：

不使用全局变量进行一次计算。

class Solution {public:    void findCombination(string tmp, const char d[10][5], string &in_data, vector<string> &rel) {        int stt = tmp.size();        if(stt == in_data.size()) {            rel.push_back(tmp);            return;        }        int di_len = strlen(d[in_data[stt] - '0']);        for(int i = 0; i < di_len; i++) {            string tmp2 = tmp;            tmp += d[in_data[stt] - '0'][i];            findCombination(tmp, d, in_data, rel);            tmp = tmp2; // d[in_data[i] - '0'][j];        }        }    vector<string> letterCombinations(string digits) {        vector<string> rel;        string tmp;        int d_len;        const char d[10][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        d_len = digits.length();        if(!d_len)            return rel;        findCombination(tmp, d, digits, rel);        return rel;    }};

http://www.cnblogs.com/longdouhzt/archive/2012/04/22/2465399.html
https://zhidao.baidu.com/question/1669208534797631547.html
http://blog.csdn.net/yunyun1886358/article/details/5659851

写法五

直接使用一个函数，而不需要分开来写。但是还是分开两个函数来做显得更加简洁。

class Solution {public:    const char d[10][5] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    string tmp;    vector<string> rel;    vector<string> letterCombinations(string digits) {        int d_len;        d_len = digits.length();        if(!d_len) {            if(tmp.empty())                return rel;            rel.push_back(tmp);            return rel;        }         int row = digits[0] - '0';        int r_len = strlen(d[row]);        for(int i = 0; i < r_len; i++) {            tmp.push_back(d[row][i]);            if(d_len > 0)                rel = letterCombinations(digits.substr(1));            tmp.pop_back();        }        return rel;    }};

class Solution {public:string num2alpha[8] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    vector<string> letterCombinations(string digits) {        vector<string> result;        if(digits.empty()) return result;        char curDigit = digits[0];        string curStr = num2alpha[curDigit-'2'];        vector<string> subResult = letterCombinations(digits.substr(1));        if(subResult.empty()) subResult.emplace_back("");        for(int i = 0; i < curStr.length(); ++i)        {            for(int j = 0; j < subResult.size(); ++j)            {                string tmp = "";                tmp += curStr[i];                tmp = tmp + subResult[j];                result.emplace_back(tmp);            }        }        return result;    }}