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.
Sample Input
23
Sample Output
ad bd cd ae be ce af bf cf
Hint
the above answer is in lexicographical order, please output in the same order!
我的做法:
import java.util.Scanner;
public class Main {
public static void main(String [] args) throws Exception {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char[] arr = str.toCharArray();
int n = arr.length;
String [][]a = new String[n][5];
int num[] = new int[n];
for(int i=0;i<arr.length;i++){
if(arr[i] == '2'){
num[i] = 3;
a[i][1]="a";
a[i][2]="b";
a[i][3]="c";
a[i][4]="*";
}
if(arr[i] == '3'){
num[i] = 3;
a[i][1]="d";
a[i][2]="e";
a[i][3]="f";
a[i][4]="*";
}
if(arr[i] == '4'){
num[i] = 3;
a[i][1]="g";
a[i][2]="h";
a[i][3]="i";
a[i][4]="*";
}
if(arr[i] == '5'){
num[i] = 3;
a[i][1]="j";
a[i][2]="k";
a[i][3]="l";
a[i][4]="*";
}
if(arr[i] == '6'){
num[i] = 3;
a[i][1]="m";
a[i][2]="n";
a[i][3]="o";
a[i][4]="*";
}
if(arr[i] == '7'){
num[i] = 4;
a[i][1]="p";
a[i][2]="q";
a[i][3]="r";
a[i][4]="s";
}
if(arr[i] == '8'){
num[i] = 3;
a[i][1]="t";
a[i][2]="u";
a[i][3]="v";
a[i][4]="*";
}
if(arr[i] == '9'){
num[i] = 4;
a[i][1]="w";
a[i][2]="x";
a[i][3]="y";
a[i][4]="z";
}
}
int number = 1;
for(int i=0;i<n;i++){
number = number*num[i];
}
//System.out.println(number);
String []out = new String[number];
for(int i=0;i<number;i++){
out[i] = "";
}
int m = 0;
for(int i=0;i<number;i++){
out[i] = out[i]+a[m][1];
i++;
out[i] = out[i]+a[m][2];
i++;
out[i] = out[i]+a[m][3];
if(a[m][4]!="*"){
i++;
out[i] = a[m][4];
}
}
m++;
while(m<n){
int k=1;
for(int i=0;i<m;i++){
if(arr[i]=='7'||arr[i]=='9'){
k = k*4;
}else {
k = k*3;
}
}
//System.out.println(k);
for(int i=0;i<number;){
for(int j=0;j<k;j++){
out[i] = out[i]+a[m][1];
i++;
}
for(int j=0;j<k;j++){
out[i] = out[i]+a[m][2];
i++;
}
for(int j=0;j<k;j++){
out[i] = out[i]+a[m][3];
i++;
}
if(a[m][4]!="*"){
for(int j=0;j<k;j++){
out[i] = out[i]+a[m][4];
i++;
}
}
}
m++;
}
for(int i=0;i<number;i++){
System.out.println(out[i]);
}
}
}
同學易浩的做法:
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
String[][] map = {{"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 [] index = new int[s.length()];
for(int i=0;i<s.length();i++){
index[i] = (s.charAt(i)-'0');
}
String[] result;
if(index.length==1){
result = map[index[0]-2];
}
else if(index.length==2){
result = fun(map[index[0]-2],map[index[1]-2]);
}
else{
int n = index.length;
result = fun(map[index[n-2]-2],map[index[n-1]-2]);
for(int i=index.length-3;i>=0;i--){
String []tmp = map[index[i]-2];
result = fun(tmp,result);
}
}
for(int i=0;i<result.length;i++){
System.out.println(result[i]);
}
}
public static String[] fun(String[] tmp,String[] tmp1){
String [] result = new String[tmp1.length*tmp.length];
for(int i=0;i<tmp1.length;i++){
for(int j=0;j<tmp.length;j++){
result[i*tmp.length+j]= tmp[j]+tmp1[i];
}
}
return result;
}
}