将基本的递归算法转换为动态的自下而上制表算法
Converting a rudimentary recursive algorithm to a dynamic bottom-up tabulation algorithm
问题陈述:
给定一个数字序列,计算给定数字序列的可能解码。
示例:
12 gives 2 for:
'AB' and 'L'
和
123 gives 3 for:
'ABC', 'LC' and 'AW'
这是我的尝试:
import java.util.*;
public class decodingcount {
static int calls = 0;
static Map<Integer, String> codes = new HashMap<Integer, String>();
private static void construct(){
codes.put(1, "A");
codes.put(2, "B");
codes.put(3, "C");
codes.put(4, "D");
//..so on
}
private static int decode(String str, String built){
construct();
int n = str.length();
int count = 0;
if (n == 0) {
System.out.println(built);
return 1;
}
// If you have 0's, then they don't have a corresponding singular letter. Break off the recursion.
if (str.substring(0, 1).equals("0"))
return 0;
String x = codes.get(Integer.parseInt(str.substring(0, 1)));
if (x == null)
return 0;
count = decode(str.substring(1), built+x);
if (n > 1) {
// If it's more than 26, it doesn't have a corresponding letter. Break off the recursion.
if (Integer.parseInt(str.substring(0, 2)) > 26)
return 0;
String y = codes.get(Integer.parseInt(str.substring(0, 2)));
count += decode(str.substring(2), built+y);
}
return count;
}
public static void main(String[] args) {
System.out.println(decode(args[0], ""));
}
}
这以指数时间运行。我真的很难将其转换为动态编程自下而上的制表算法。 Here is my attempt
.它 returns 0。非常感谢任何帮助。
工作代码:
private static int decode(String str) {
construct();
int n = str.length();
int[] count = new int[n];
count[0] = 1;
for (int i = 1; i < n; i++) {
String x = codes.get(Integer.parseInt(str.substring(i, i + 1)));
if (str.charAt(i) != '0')
count[i] = count[i - 1];
if (Integer.parseInt(str.substring(i - 1, i + 1)) <= 26 && str.charAt(i - 1) != '0')
count[i] += (i - 2 >= 0 ? count[i - 2] : 1);
}
return count[n - 1];
}
问题陈述: 给定一个数字序列,计算给定数字序列的可能解码。
示例:
12 gives 2 for:
'AB' and 'L'
和
123 gives 3 for:
'ABC', 'LC' and 'AW'
这是我的尝试:
import java.util.*;
public class decodingcount {
static int calls = 0;
static Map<Integer, String> codes = new HashMap<Integer, String>();
private static void construct(){
codes.put(1, "A");
codes.put(2, "B");
codes.put(3, "C");
codes.put(4, "D");
//..so on
}
private static int decode(String str, String built){
construct();
int n = str.length();
int count = 0;
if (n == 0) {
System.out.println(built);
return 1;
}
// If you have 0's, then they don't have a corresponding singular letter. Break off the recursion.
if (str.substring(0, 1).equals("0"))
return 0;
String x = codes.get(Integer.parseInt(str.substring(0, 1)));
if (x == null)
return 0;
count = decode(str.substring(1), built+x);
if (n > 1) {
// If it's more than 26, it doesn't have a corresponding letter. Break off the recursion.
if (Integer.parseInt(str.substring(0, 2)) > 26)
return 0;
String y = codes.get(Integer.parseInt(str.substring(0, 2)));
count += decode(str.substring(2), built+y);
}
return count;
}
public static void main(String[] args) {
System.out.println(decode(args[0], ""));
}
}
这以指数时间运行。我真的很难将其转换为动态编程自下而上的制表算法。 Here is my attempt .它 returns 0。非常感谢任何帮助。
工作代码:
private static int decode(String str) {
construct();
int n = str.length();
int[] count = new int[n];
count[0] = 1;
for (int i = 1; i < n; i++) {
String x = codes.get(Integer.parseInt(str.substring(i, i + 1)));
if (str.charAt(i) != '0')
count[i] = count[i - 1];
if (Integer.parseInt(str.substring(i - 1, i + 1)) <= 26 && str.charAt(i - 1) != '0')
count[i] += (i - 2 >= 0 ? count[i - 2] : 1);
}
return count[n - 1];
}