17. Letter Combinations of a Phone Number题解
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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"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
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题意:
给一个数字的字符串,输出对应键盘上的所有字符的组合。
分析:
对于数字的字符串23来说,可以先生成一个List<String> result =new ArrayList<>()保存结果。
(Java中List<String> result =new ArrayList<>();List<String> result =new ArrayList<>();List是集合最大的父类,它包含了ArrayList。 如果直接声明为ArrayList<String> list=new ArrayList<String>()这个也没有问题。 而声明成:List<String> list=new ArrayList<String>();这样的形式使得list这个对象可以有多种的存在形式,比如要用链表存数据的话直接用LinkedList,使用ArrayList或者Vector直接通过list去=就可以了,这样让list这个对象活起来了)
对字符串的每个数字进行遍历,首先为2,则把‘2‘-’0‘对应的字符分别加入到res临时List中,此时res为{’a‘,'b','c'};
令result = res; 字符串此时指向3,这是把’3‘-’0‘对应的每个字符,分别加入到result目前已有的字符的后面。
即一个双重循环遍历。
源代码:
public static List<String> letterCombinations(String digits) {
List<String> result =new ArrayList<>();
if(digits == null || digits.isEmpty())
return result;
result.add("");
String []btns = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
for(int i =0 ; i < digits.length() ;i++)
{
List<String> res = new ArrayList<>();
String letter = btns[digits.charAt(i)-'0'];
for(int j = 0 ; j < result.size();j++)//遍历上一个列表,取出每一个元素,并和新的元素的每一个字符加起来保存
{
for(int k = 0; k< letter.length(); k++)//遍历当前数字对应的所有字符
{
res.add(result.get(j)+letter.charAt(k));
}
}
result = res;
}
return result;
}
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