Letter Combinations of a Phone Number
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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"].
解答:
我的答案:回溯,时间慢 2ms
class Solution {private:vector<string> vec{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};int number = 0;vector<string> vs;string digit;public: void combine(string str, int l){ if(l == number){ vs.push_back(str); return; } int nums = digit[l]-'0'; string alpha = vec[nums]; for(int k = 0; k < alpha.size(); k++){ combine(str+alpha[k],l+1); } } vector<string> letterCombinations(string digits) { int len = digits.size(); string s; if(len < 1) return vs; number = len; digit = digits; combine(s,0); return vs; }};
2.别人的代码 快,0ms
对别人优秀的代码致敬!
vector<string> letterCombinations(string digits) { vector<string> result; if(digits.empty()) return vector<string>(); static const vector<string> v = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; result.push_back(""); // add a seed for the initial case for(int i = 0 ; i < digits.size(); ++i) { int num = digits[i]-'0'; if(num < 0 || num > 9) break; const string& candidate = v[num]; if(candidate.empty()) continue; vector<string> tmp; for(int j = 0 ; j < candidate.size() ; ++j) { for(int k = 0 ; k < result.size() ; ++k) { tmp.push_back(result[k] + candidate[j]); } } result.swap(tmp); } return result;}
Explanation with sample input "123"
Initial state:
- result = {""}
Stage 1 for number "1":
- result has {""}
- candiate is "abc"
- generate three strings "" + "a", ""+"b", ""+"c" and put into tmp, tmp = {"a", "b","c"}
- swap result and tmp (swap does not take memory copy)
- Now result has {"a", "b", "c"}
Stage 2 for number "2":
- result has {"a", "b", "c"}
- candidate is "def"
- generate nine strings and put into tmp, "a" + "d", "a"+"e", "a"+"f", "b" + "d", "b"+"e", "b"+"f", "c" + "d", "c"+"e", "c"+"f"
- so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
- swap result and tmp
- Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
Stage 3 for number "3":
- result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
- candidate is "ghi"
- generate 27 strings and put into tmp,
- add "g" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- so, tmp has {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg" "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh" "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
- swap result and tmp
- Now result has {"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg" "adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh" "adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
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