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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"].
/|\ /|\ /|\"425"|g h i 4
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 | class Solution {public: vector<string> alphabets = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; vector<string> letterCombinations(string digits) { int len = digits.size(); vector<string> result; if( len == 0) return result; dfs(digits, 0, "", result); return result; } void dfs(string digits, int pos, string path, vector<string> & result){ if(digits.size() == pos){ result.push_back(path); return; } for(auto d: alphabets[digits[pos] - ‘0‘]){ dfs(digits, pos + 1, path + d, result); } }}; |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | class Solution {public: vector<string> charmap = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; vector<string> letterCombinations(string digits) { vector<string> res; if(digits.size() == 0) return res; 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; //重新赋值结果vector数组 } return res; }}; |
17. Letter Combinations of a Phone Number
标签：... ext com 方法 alpha turn osi 搜索 tom
原文地址：http://www.cnblogs.com/zhxshseu/p/b69a3b31deb4cdab42dd8d2ca22e1ca7.html
