Given a string contains only numeric characters 2-9, it can return all letter combinations indicated.
Given digital map to letters as follows (the same telephone key). Note 1 does not correspond to any alphabet.
Example: Enter "23"
Output: [ad, bd, cd, ad, ae, af, bd, be, bf, cd, ce, cf]
Analysis:
First, a switch case statement numbers and the numbers corresponding to letters out .
Then give a fixed string. Each time you add a letter to the inside, before adding too as a whole, the first group of new coverage before, because there are duplicate.
Code:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class pipeiSolution {
public static String mapping(char n){ //数字到字母的对应
String str=null;
switch(n){
case '0':
str="";
break;
case'1':
str="";
break;
case'2':
str="abc";
break;
case'3':
str="def";
break;
case'4':
str="ghi";
break;
case'5':
str="jkl";
break;
case'6':
str="mno";
break;
case'7':
str="pqrs";
break;
case'8':
str="tuv";
break;
case'9':
str="wxyz";
break;
}
return str;
}
public static List<String> letterCombinations(String datas){
List<String> ret=new ArrayList<>();
int i=0;
for(i = 0;i<datas.length();i++){
char c=datas.charAt(i);
int length=ret.size();
String substring=mapping(c);
if(length==0){
for(int j=0;j<substring.length();j++){
ret.add(" "+substring.charAt(j)); //直接加
}
}
else{
for(int p=0;p<length;p++){
for(int m=0;m<substring.length();m++){
ret.add(ret.get(p)+substring.charAt(m)); //加入除第一组的其他组合
}
ret.set(p,(ret.get(p)+substring.charAt(0))); //覆盖之前的
}
}
}
return ret;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println("please input the string: ");
String datas=scan.nextLine();
System.out.println(letterCombinations(datas));
}
}
operation result:
