将两个十进制整数与除法运算相乘
我试着将两个数相乘,它们是正整数,它们的位数相同,通过递归的除法和征服,我试着这样做:T(n)=4T(n/2)+O(n) 注意:我知道它以θ(n^2)运行,这很糟糕!这对我来说只是一种锻炼。 谢谢你,很抱歉我的英语不好。:) 我的问题是:我的错误在哪里? algorithm based on this doc 以下是代码:
import java.util.Scanner;
public class main {
static int res=0;
static int stage =0;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
char[] Num1;
char[] Num2;
String num1 = in.nextLine();
String num2 = in.nextLine();
in.close();
Num1 = num1.toCharArray();
Num2 = num2.toCharArray();
DaQMultiplay(Num1, Num2);
System.out.println(res);
}
static int DaQMultiplay(char[] num1,char[] num2){
if(num1.length<2){
stage++;
int num1sd =Integer.parseInt(new String(num1));
int num2sd =Integer.parseInt(new String(num2));
return (num1sd*num2sd);
}
stage++;
double len = num1.length;
int lenl = (int) Math.ceil(len/2);
char []ln1 = new char[lenl];
char []rn1 = new char[(int) (len-lenl)];
char []ln2 = new char[lenl];
char []rn2 = new char[(int) (len-lenl)];
for (int i = 0; i < ln1.length; i++) {
ln1[i]=num1[i];
}
for (int i = 0; i < rn1.length; i++) {
rn1[i]=num1[i+lenl];
}
for (int i = 0; i < ln2.length; i++) {
ln2[i]=num2[i];
}
for (int i = 0; i < rn2.length; i++) {
rn2[i]=num2[i+lenl];
}
System.out.print("Left Side of num1:"+stage+" ");
System.out.println(ln1);
System.out.print("Right Side of num1:"+stage+" ");
System.out.println(rn1);
System.out.print("Left Side of num2:"+stage+" ");
System.out.println(ln2);
System.out.print("Right Side of num2:"+stage+" ");
System.out.println(rn2);
res+=DaQMultiplay(ln1,ln2)*(10^((int)len));
System.out.println("res: "+res);
res+=DaQMultiplay(ln1,rn2)*10^((int) (len-lenl));
System.out.println("res: "+res);
res+=DaQMultiplay(rn1,ln2)*10^((int) (len-lenl));
System.out.println("res: "+res);
res+=DaQMultiplay(rn1, rn2);
System.out.println("res: "+res);
return 0;
}
}
输出:对于num1=20011,num2=91281
20011
91281
Left Side of num1:1 200
Right Side of num1:1 11
Left Side of num2:1 912
Right Side of num2:1 81
Left Side of num1:2 20
Right Side of num1:2 0
Left Side of num2:2 91
Right Side of num2:2 2
Left Side of num1:3 2
Right Side of num1:3 0
Left Side of num2:3 9
Right Side of num2:3 1
res: 144
res: 164
res: 164
res: 0
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
at main.DaQMultiplay(main.java:46)
at main.DaQMultiplay(main.java:63)
at main.DaQMultiplay(main.java:61)
at main.main(main.java:19)
# 1 楼答案
通常,您的代码不会处理num2解析为num1之前的一位数的情况。这将导致通过DaQ方法生成一个空字符串,最终引发异常。首先需要添加对num2解析处理的检查。该检查解决了第一个异常(第46行附近):
然后你需要在乘法阶段加一个检查: int num1sd=1; int num2sd=1
我不确定第二个检查是否适用于您编写的算法,但总体思路是,这个if语句
if(num1.length<2){...
只适用于num1首先解析的情况,并不总是这样更正了代码,但仍然传递了错误的答案:
新输出:num1=,num2=
算法的实现问题仍然存在
# 2 楼答案
以下是完整的程序,一切正常: