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17. Letter Combinations of a Phone Number
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"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
BFS:
1 class Solution { 2 public: 3 vector<string> letterCombinations(string digits) 4 { 5 vector<string> vs = {"", "", "abc", "def","ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" }; 6 vector<string> result; 7 result.push_back(""); 8 if(digits.empty()) return {}; 9 for (size_t i = 0; i < digits.size(); ++i) { 10 vector<string> qc; 11 string strtmp(vs[digits[i]-‘0‘]); 12 for (size_t j = 0; j < strtmp.size(); ++j) { 13 for (size_t k = 0; k < result.size(); ++k) { 14 qc.push_back(result[k] + strtmp[j]); 15 } 16 } 17 result = qc; 18 } 19 return result; 20 } 21 };
2% 3ms
对于含有0和1的项,比如"123",程序运行结果不符合预期.但是却通过OJ,所以此题有待严谨.
BFS:
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> vs = {"", "", "abc", "def","ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" }; vector<string> result; result.push_back(""); if(digits.empty()) return {}; for (size_t i = 0; i < digits.size(); ++i) { vector<string> qc; string strtmp(vs[digits[i]-‘0‘]); for (size_t j = 0; j < strtmp.size(); ++j) { for (size_t k = 0; k < result.size(); ++k) { qc.push_back(result[k] + strtmp[j]); } } result.swap(qc);//swap does not take memory copy } return result; } };
the swap method thought by asbear
65.33% 0ms
backtrackiing:
1 class Solution { //by luming.zhang.75(Creater) redace85(Modifier) 2 public: 3 vector<string> letterCombinations(string digits) { 4 vector<string> ret; 5 if(0>=digits.size()) return ret; 6 7 const string map[]={"0","1","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; 8 backTrackingFun(map,digits,"",ret); 9 10 return ret; 11 } 12 13 void backTrackingFun(const string m[], const string &digits, string r, vector<string> &ret){ 14 int c=r.size(); 15 if(digits.size()==c){ 16 ret.push_back(r); 17 return; 18 } 19 20 auto str = m[digits.at(c)-48]; 21 for(auto it=str.cbegin();it!=str.cend();++it){ 22 r.push_back(*it); 23 backTrackingFun(m,digits,r,ret); 24 r.pop_back(); 25 } 26 } 27 };
2% 3ms
17. Letter Combinations of a Phone Number
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