LeetCode 17. Letter Combinations of a Phone Number
https://leetcode.com/problems/letter-combinations-of-a-phone-number/description/
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.
- 字符串组合 + FIFO queue
- 两种解法，一种用到queue，另一种用vector模拟queue使用。注意初始化比较tricky的地方。
- queue - C++ Reference
- http://www.cplusplus.com/reference/queue/queue/
- vector::swap - C++ Reference
- http://www.cplusplus.com/reference/vector/vector/swap/
1 // 2 // main.cpp 3 // LeetCode 4 // 5 // Created by Hao on 2017/3/16. 6 // Copyright © 2017年 Hao. All rights reserved. 7 // 8 // 9 // main.cpp 10 // LeetCode 11 // 12 // Created by Hao on 2017/3/16. 13 // Copyright © 2017年 Hao. All rights reserved. 14 // 15 #include <iostream> 16 #include <string> 17 #include <vector> 18 #include <queue> 19 using namespace std; 20 21 class Solution { 22 public: 23 vector<string> letterCombinations(string digits) { 24 vector<string> vResult; 25 26 if (digits.empty()) return vResult; 27 28 vector<string> dict{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; 29 30 queue<string> myQueue; 31 myQueue.push(""); // Tricky step for the initial value 32 33 for (int i = 0; i < digits.size(); i ++) { 34 int n = myQueue.size(); 35 string sTemp; 36 37 for (int j = 0; j < n; j ++) { 38 // retrieve old letter 39 sTemp = myQueue.front(); 40 myQueue.pop(); 41 42 // append new letter 43 for (auto & ch : dict[digits.at(i) - '0']) 44 myQueue.push(sTemp + ch); 45 } 46 } 47 48 while (! myQueue.empty()) { 49 vResult.push_back(myQueue.front()); 50 myQueue.pop(); 51 } 52 53 return vResult; 54 } 55 56 vector<string> letterCombinations2(string digits) { 57 vector<string> vResult; 58 59 if (digits.empty()) return vResult; 60 61 vector<string> dict{"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; 62 63 vResult.push_back(""); // Tricky step for the initial value 64 65 for (int i = 0; i < digits.size(); i ++) { 66 vector<string> vTemp; // temp result to override the initial value "" 67 string sTemp; 68 69 for (int j = 0; j < vResult.size(); j ++) { 70 sTemp = vResult[j]; 71 72 for (auto & ch : dict[digits.at(i) - '0']) 73 vTemp.push_back(sTemp + ch); 74 } 75 76 vResult.swap(vTemp); // update result iteratively 77 } 78 79 return vResult; 80 } 81 }; 82 83 int main(int argc, char* argv[]) 84 { 85 Solution mySolution; 86 87 vector<string> inputs {"", "23"}; 88 vector<string> outputs; 89 90 /* 91 92 ad ae af bd be bf cd ce cf 93 */ 94 for (auto & input : inputs) { 95 outputs = mySolution.letterCombinations(input); 96 97 for (auto & output : outputs) { 98 cout << output << " "; 99 } 100 cout << endl; 101 102 outputs = mySolution.letterCombinations2(input); 103 104 for (auto & output : outputs) { 105 cout << output << " "; 106 } 107 cout << endl; 108 } 109 110 return 0; 111 }
- 回溯法
- 学习Solution里Pythonic写法
1 class Solution: 2 def letterCombinations(self, digits: str) -> List[str]: 3 if not digits: 4 return [] 5 6 self._mapping = ['', '', 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'] 7 results = [] 8 self.helper(results, digits, '', 0) 9 10 return results 11 12 def helper(self, results: list, digits: str, current: str, n: int): 13 if n == len(digits): 14 results.append(current) 15 else: 16 for ch in self._mapping[int(digits[n])]: 17 self.helper(results, digits, current + ch, n + 1)
