leetcode-17-letter-combinations-of-a-phone-number
给定一个数字字符串，找出这个字符串对应的字母串组合,这是一个简单题，练习一下linkedlist作为队列的用法
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.
- 数字与字母的对应关系同电话键盘，2–>abc 3–>def…
保存对应关系，遍历数字串，在结果集中追加当前数字对应字母. 我怎么没想到用队列…
- 我的代码
class Solution {
public List<String> letterCombinations(String digits) {
List<String> res = new ArrayList<String>();
if(digits==null || "".equals(digits))
return res;
String dic [][] = {
{},
{},
{"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"}
};
int len = digits.length();
int t = (int)(digits.charAt(0)-'0');
for(int i = 0; i < dic[t].length; i++){
res.add(dic[t][i]);
}
for(int j = 1; j < len; j++){
int n = (int)(digits.charAt(j)-'0');
List<String> tempres = new ArrayList<String>();
for(int i= 0; i < dic[n].length; i++){
for(String s :res){
tempres.add(s+dic[n][i]);
}
}
res = tempres;
tempres = null;
}
return res;
}
}
- 题解代码
public List<String> letterCombinations(String digits) {
LinkedList<String> ans = new LinkedList<String>();
String[] mapping = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
ans.add("");
for(int i =0; i<digits.length();i++){
int x = Character.getNumericValue(digits.charAt(i));
while(ans.peek().length()==i){
String t = ans.remove();
for(char s : mapping[x].toCharArray())
ans.add(t+s);
}
}
return ans;
}
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