[LeetCode][Java] Letter Combinations of a Phone Number
题目: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
题目:
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.
题意:
给定一个数字字符串,返回这个数字能代表的所有的字母的组合。
数字对字母的映射如下图所示(就在电话的按键中)
Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].返回的结果可以按照任意的顺序。
算法分析:
* 思路:
*比如“234”这个字符串,我可以先将0...1的所有排列找到-->{"a", "b", "c"}
*再进一步将0...2的所有排列找到-->{"ad", "ae","af", "bd", "be", "bf", "cd", "ce", "cf"}
*如此循环...直到字符串末尾。实现如下
AC代码:
public class Solution
{
public ArrayList<String> letterCombinations(String digits)
{
ArrayList<String> res=new ArrayList<String>();
String charmap[] = {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
if (digits.length()==0) return res;
res.add("");
for (int i = 0; i < digits.length(); i++)
{
ArrayList<String> tempres=new ArrayList<String>();
String chars = charmap[digits.charAt(i) - '0'];
for (int c = 0; c < chars.length();c++)
{
for (int j = 0; j < res.size();j++)
tempres.add(res.get(j)+chars.charAt(c));
}
res = tempres;
}
return res;
}
}
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