Letter combinations of a phone number [leetcode] Two ideas about the cyclic Solution

There are many two ideas in this series of blog posts, because there is an idea when I click questions for the first time, and I forget my previous ideas when I click questions for the second time, there are new ideas.

At the same time, I also Po most of the Code to the Q & A of the corresponding question of leetcode, So if you also look at the problem discussion, you will find the same code as me, actually my PO :)

The following two methods are used to connect books to the text:

First approach

For example, for the string "234", I can first locate all the arrays of 0... 1 --> {"A", "B", "C "}

Then add 0... for all the arrays of 2, go to --> {"ad", "AE", "af", "BD", "be", "BF", "cd", "CE ", "CF "}

So loop... until the end of the string. The implementation is as follows:

vector<string> letterCombinations(string digits) {        vector<string> res;        string charmap[10] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};        res.push_back("");        for (int i = 0; i < digits.size(); i++)        {            vector<string> tempres;            string chars = charmap[digits[i] - '0'];            for (int c = 0; c < chars.size();c++)                for (int j = 0; j < res.size();j++)                    tempres.push_back(res[j]+chars[c]);            res = tempres;        }        return res;    }

Second approach

Mr. Cheng is the first feasible solution "ADG" of "234", and other solutions are derived from the feasible solution.

When the number is similar to the number, 9 is changed to 0 and carry.

Vector <string> lettercombinations (string digits) {string charmap [10] = {"0", "1", "ABC", "def", "Ghi ", "jkl", "MnO", "pqrs", "TUV", "wxyz"}; vector <string> res; string Alpha; For (INT I = 0; I <digits. size (); I ++) Alpha + = charmap [digits [I]-'0'] [0]; while (true) {res. push_back (alpha); // find the feasible solution bool find = false; For (INT I = digits. size ()-1; I >=- 1 &&! Find; I --) {if (I =-1) return res; // The End Of The traversal string chars = charmap [digits [I]-'0']; if (alpha [I] = chars [chars. size ()-1]) // traverses the last feasible solution of the number I, reset and find the feasible solution of the number I + 1 {Alpha [I] = chars [0]; continue;} For (INT C = 0; C <chars. size ()&&! Find; C ++) // traverse other feasible solutions of the number I {If (alpha [I] = chars [c]) {Alpha [I] = chars [C + 1]; find = true ;}}}}}

Letter combinations of a phone number [leetcode] Two ideas about the cyclic Solution