17. Letter Combinations of a Phone Number
Problem:
Given a string containing digits from 2-9
inclusive, 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. Note that 1 does not map to any letters.
Example:
Input: "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.
Solution:
ONE
JavaScript specific optimization.
Array.prototype.push
accepts arbitrary arguments which enables tighter loops.
Also, appending string is faster than prepending.
///**
* @param {string} digits
* @return {string[]}
*/let letterCombinations = function (digits) {if (digits.length <= 0) {return [];}const letters = [,,['a', 'b', 'c'],['d', 'e', 'f'],['g', 'h', 'i'],['j', 'k', 'l'],['m', 'n', 'o'],['p', 'q', 'r', 's'],['t', 'u', 'v'],['w', 'x', 'y', 'z']];let result = [''];for (let i = 0; i < digits.length; i++) {const arr = letters[digits[i]];let newResult = [];
arr.forEach((c) => newResult.push(...result.map((r) => r + c)));
result = newResult;}return result;};
TWO
General recursive DFS solution.
///**
* @param {string} digits
* @return {string[]}
*/let letterCombinations = function (digits) {const letters = [, , 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];const result = [];if (digits.length > 0) {dfs(digits, 0, '', letters, result);}return result;};function dfs(digits, idigit, path, letters, result) {if (idigit >= digits.length) {
result.push(path);return;}const str = letters[digits[idigit]];for (let i = 0; i < str.length; i++) {dfs(digits, idigit + 1, path + str[i], letters, result);}}
☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆
☆: .。. o(≧▽≦)o .。.:☆☆: .。. o(≧▽≦)o .。.:☆