这是我自己的方法,我也在使用我自己的BigInt课程,我做了所有事情,但如果有人可以帮助我,我似乎无法找出我做错的地方,我会非常感激。
public BigInt mul(BigInt o) {
int max = n.length > o.n.length ? n.length : o.n.length;
int[] newdigits = new int[n.length + o.n.length];
for (int i = 0; i < max; i++)
{
int carry = 0;
for (int i2 = 0; i2 < o.n.length || carry > 0; i2++)
{
int otherDigit = i2 >= o.n.length ? 0: o.n[i2];
int val = (n[i] * otherDigit) + carry;
newdigits[i + i2] += val % 10;
carry = val / 10;
}
}
return new BigInt(newdigits);
}
这是我的代码到目前为止它可以工作,但在结果中我得到一个额外的零,作为一个例子:当我乘以100*10
我得到01000
而不是1000
和其他问题就是当我乘以9999*1999
时,我得到了这个:1817262619101
,但正确的答案是19988001
。
有人可以帮我这个吗?
这是我的BigInt课程:
public class BigInt {
public static void main(String[] args) {
BigInt a = new BigInt(args.length == 2 ? args[0] : "9999");
BigInt b = new BigInt(args.length == 2 ? args[1] : "1999");
System.out.println(a + (a.equals(b) ? " equals " : " does not equal ") + b);
System.out.println(a + (a.compareTo(b) < 0 ? " < " : (a.compareTo(b) > 0 ? " > " : " = ")) + b);
System.out.println(a + " + " + b + " = " + a.add(b));
if (a.compareTo(b) >= 0) {
System.out.println(a + " - " + b + " = " + a.sub(b));
}
System.out.println(a + " * " + b + " = " + a.mul(b));
if (a.compareTo(b) >= 0) {
System.out.println(a + " / " + b + " = " + a.div(b));
}
}
}
final class BigInt implements Comparable<BigInt> {
int[] digits;
int size;
public BigInt() {
n = new int[1];
}
public BigInt(String s) {
n = new int[s.length()];
for (int i = 0; i < n.length; ++i) {
n[n.length - i - 1] = s.charAt(i) - '0';
}
n = trim(n);
}
private BigInt(int[] n) {
this.n = new int[n.length];
for (int i = 0; i < n.length; ++i) {
this.n[i] = n[i];
}
}
public BigInt add(BigInt o) {
return null;
}
public int compareTo(BigInt o) {
if (n.length < o.n.length) {
return -1;
}
else if (n.length > o.n.length) {
return +1;
}
for (int i = n.length-1; i >= 0; --i) {
if (n[i] < o.n[i]) {
return -1;
}
else if (n[i] > o.n[i]) {
return +1;
}
}
return 0;
}
public BigInt div(BigInt o) {
return null;
}
public boolean equals(Object o) {
if (o instanceof BigInt) {
if (n.length == ((BigInt)o).n.length) {
for (int i = 0; i < n.length; ++i) {
if (n[i] != ((BigInt)o).n[i]) {
return false;
}
}
return true;
}
}
return false;
}
public BigInt mul(BigInt o) {
int max = n.length > o.n.length ? n.length : o.n.length;
int[] newdigits = new int[n.length + o.n.length];
for (int i = 0; i < max; i++)
{
int carry = 0;
for (int i2 = 0; i2 < o.n.length || carry > 0; i2++)
{
int otherDigit = i2 >= o.n.length ? 0: o.n[i2];
int val = (n[i] * otherDigit) + carry;
newdigits[i + i2] += val % 10;
carry = val / 10;
}
}
return new BigInt(newdigits);
}
public BigInt sub(BigInt o) {
return null;
}
public String toString() {
String s = "";
for (int i : n) {
s = i + s;
}
return s;
}
private int[] trim(int[] nums) {
int size = nums.length;
for (int i = nums.length - 1; i > 0; --i) {
if (nums[i] != 0) {
break;
}
--size;
}
int[] res = new int[size];
for (int i = 0; i < size; ++i) {
res[i] = nums[i];
}
return res;
}
private int[] n;
}
答案 0 :(得分:6)
没有尝试调试代码,但这就是我实现这种乘法的方法。效率稍低,但更容易跟踪进位。
public BigInt mul(BigInt o) {
int max = n.length > o.n.length ? n.length : o.n.length;
int[] newdigits = new int[n.length + o.n.length];
for (int i = 0; i < max; i++) {
for (int i2 = 0; i2 < max; i2++) {
int digit1 = i >= n.length ? 0 : n[i];
int digit2 = i2 >= o.n.length ? 0 : o.n[i2];
if (digit1 > 0 && digit2 > 0) {
int value = digit1 * digit2;
int pos = i + i2;
while (value > 0) {
int newDigit = (newdigits[pos] + value) % 10;
value = (newdigits[pos] + value) / 10;
newdigits[pos] = newDigit;
pos++;
}
}
}
}
return new BigInt(newdigits);
}
答案 1 :(得分:3)
首先,查看可能的长度:
[1] * [1] = [1] OR长度(1)+长度(1) - &gt;长度(l)
[9] * [9] = [8,1] OR长度(1)+长度(1) - >长度(2)
一种解决方案
Psudocode:
if (newdigits[ (length-1) ] == 0)
int[] newarray = new int[ (length-1) ]
copy newdigits to newarray
return newarray
你必须查看前面数字之前的最大数字的进位。
其次,您需要将找到的先前值添加到val变量中。
i = 0结束:
[1,9,9,7,1,0,0,0]
在i = 1期间添加:
[0,1,9,9,7,1,0,0]
i = 1的结尾:
[1,10,18,16,8,1,0,0]
我建议您尝试使用Arrays.toString(newdigits)或使用调试器来捕获这些问题。