Letter Combinations of a Phone Number (M)
題目
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.
題意
給定一個只包含 2-9 數字的字串,要求回傳由這些數字代表的字母組成的所有排列,
思路
遞回回溯法或迭代實作,
代碼實作
Java
遞回
class Solution {
public List<String> letterCombinations(String digits) {
// 特殊情況,直接回傳空的List
if (digits.isEmpty()) {
return new ArrayList<String>();
}
String[] map = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
List<String> ans = new ArrayList<>();
generate(digits, map, "", ans);
return ans;
}
private void generate(String digits, String[] map, String s, List<String> ans) {
// 遞回邊界,無剩余數字可用說明已經完成了一個排列
if (digits.isEmpty()) {
ans.add(s);
return;
}
int first = digits.charAt(0) - '0';
for (int i = 0; i < map[first].length(); i++) {
s += map[first].charAt(i); // 先加入當前字符
generate(digits.substring(1), map, s, ans); // 遞回處理右側數字
s = s.substring(0, s.length() - 1); // 洗掉加入的字符
}
}
}
迭代
class Solution {
public List<String> letterCombinations(String digits) {
if (digits.isEmpty()) {
return new ArrayList<>();
}
String[] map = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };
List<String> list = new ArrayList<>();
list.add("");
for (int i = 0; i < digits.length(); i++) {
List<String> temp = new ArrayList<>();
for (char c : map[digits.charAt(i) - '0'].toCharArray()) {
for (String s : list) {
temp.add(s + c);
}
}
list = temp;
}
return list;
}
}
JavaScript
遞回
/**
* @param {string} digits
* @return {string[]}
*/
var letterCombinations = function (digits) {
if (digits.length === 0) return []
let map = ['', '', 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
let list = []
dfs(digits.split(''), 0, map, [], list)
return list
}
let dfs = function (digits, index, map, s, list) {
if (index === digits.length) {
return list.push(s.join(''))
}
for (let c of map[+digits[index]].split('')) {
s.push(c)
dfs(digits, index + 1, map, s, list)
s.pop()
}
}
迭代
/**
* @param {string} digits
* @return {string[]}
*/
var letterCombinations = function (digits) {
if (digits.length === 0) return []
let map = ['', '', 'abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
let list = ['']
for (let d of digits) {
let temp = []
for (let s of list) {
for (let c of map[+d].split('')) {
temp.push(s + c)
}
}
list = temp
}
return list
}
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