用分而治之java乘以两个十进制整数答案

【问题标题】：Multiplying two decimal integer with divide and conquer java用分而治之java乘以两个十进制整数
【发布时间】：2017-06-01 02:36:35
【问题描述】：

我正在尝试将两个数字相乘，它们是正整数并且它们具有相同的位数，递归地分而治之，我正在尝试这样做：T(n)=4T(n /2)+O(n) 注意：我知道它在 theta(n^2) 中运行，这很糟糕！这对我来说只是一个练习。谢谢你，对不起我的英语不好。 :) 我的问题是：我的错误在哪里？ algorithm based on this doc 这是代码：

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)

【问题讨论】：

标签： java divide-and-conquer multiplication

【解决方案1】：

通常，您的代码不会处理 num2 在 num1 之前解析为单个数字的情况。这会通过 DaQ 方法生成一个空字符串，最终引发异常。您需要首先添加对处理 num2 解析的检查。此检查解决了第一个异常（大约第 46 行）：

 for (int i = 0; i < rn2.length; i++) {
    if(num2.length>i+lenl){
        rn2[i]=num2[i+lenl];
    }
  }

然后你需要在乘法阶段添加一个检查： int num1sd = 1; int num2sd = 1;

if(num1!=null && !num1.equals("") && new String(num2).trim().length()>0){
    num1sd =Integer.parseInt(new String(num1));
}

if(num2!=null && !num2.equals("") && new String(num2).trim().length()>0){
    num2sd=Integer.parseInt(new String(num2));
}

我不确定第二次检查是否适合您的算法，但总体思路是这个 if 语句 if(num1.length<2){...only 处理 num1 首先解析的情况，但情况并非总是如此。

更正了代码，但它仍然传递了错误的答案：

import java.util.Scanner;


public class main {
static int res=0;
static int pow;
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();
    pow = Num1.length;



    DaQMultiplay(Num1, Num2);
    System.out.println(res);
}
static int DaQMultiplay(char[] num1,char[] num2){
    if(num1.length<2){
        stage++;
        int num1sd = 0;
        int num2sd = 0;
        if(num1!=null && !num1.equals("") && new String(num2).trim().length()>0){
            num1sd =Integer.parseInt(new String(num1));
        }

        if(num2!=null && !num2.equals("") && new String(num2).trim().length()>0){
            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++) {
        if(num2.length>i+lenl){
            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)*(Math.pow(10, len)));
    System.out.println(res);
    res+=(DaQMultiplay(rn1,ln2)*(Math.pow(10, (len/2))));
    System.out.println(res);
    res+=(DaQMultiplay(ln1,rn2)*(Math.pow(10, (len/2))));
    System.out.println(res);
    res+=(DaQMultiplay(rn1, rn2));
    System.out.println(res);
    return 0;
}
}

新输出：num1=,num2=

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
1800
1800
1820
1820
0
0
Left Side of num1:9 2
Right Side of num1:9 0
Left Side of num2:9 2
Right Side of num2:9

而且算法的实现问题依然存在……

【讨论】：

谢谢，它解决了错误，但在征服步骤中算法的正确性仍然存在问题，我不知道如何处理 4 个传入的递归尾部，我的算法基于此文档：@ 987654321@

【解决方案2】：

这是完整的程序，一切正常：

import java.io.ObjectInputStream.GetField;
import java.util.Scanner;

public class main {
    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();  
        check(Num1, Num2);
    }
    static int DaQMultiplay(char[] num1,char[] num2){
        int res=0;
        if(num1.length<=1){
            int num1sd = 0;
            int num2sd = 0;
            if(num1!=null && !num1.equals("") && new String(num2).trim().length()>0){
                num1sd =Integer.parseInt(new String(num1));
            }

            if(num2!=null && !num2.equals("") && new String(num2).trim().length()>0){
                num2sd=Integer.parseInt(new String(num2));
            }
            return (num1sd*num2sd);
        }
        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++) {
            if(num2.length>i+lenl){
                rn2[i]=num2[i+lenl];
            }
        }

        res+=(DaQMultiplay(ln1,ln2)*(Math.pow(10, len)));
        res+=(DaQMultiplay(rn1,ln2)*(Math.pow(10, (len/2))));
        res+=(DaQMultiplay(ln1,rn2)*(Math.pow(10, (len/2))));
        res+=(DaQMultiplay(rn1, rn2));

        return res;
    }
    static void check(char []Num1,char []Num2){
        int res=0;
        if ((Num1.length%2)==0) {
            res=DaQMultiplay(Num1, Num2);
            System.out.println(res);
        }
        else{
            String num1 = String.valueOf(Num1);
            num1 = num1+"0";
            String num2 = String.valueOf(Num2);
            num2 = num2+"0";
            Num1=num1.toCharArray();
            Num2=num2.toCharArray();
            res=DaQMultiplay(Num1, Num2);
            res=(res/100);
            System.out.println(res);
        }
    }
}

【讨论】：