[英]identify prime number using recursive method [java]
the problem is below.问题在下面。 main() checks numbers 1-10 by calling isPrime().
main() 通过调用 isPrime() 检查数字 1-10。 I think I have the math right however every number other than 2 comes back as not prime.
我想我的数学是对的,但是除了 2 以外的每个数字都不是素数。
I have checked some of the solutions and questions on SO, however, I can't seem to achieve the same results.我已经检查了一些关于 SO 的解决方案和问题,但是,我似乎无法获得相同的结果。
original problem:原问题:
public class PrimeChecker {
// Returns 0 if value is not prime, 1 if value is prime
public static int isPrime(int testVal, int divVal) {
// Base case 1: 0 and 1 are not prime, testVal is not prime
// Base case 2: testVal only divisible by 1, testVal is prime
// Recursive Case
// Check if testVal can be evenly divided by divVal
// Hint: use the % operator
// If not, recursive call to isPrime with testVal and (divVal - 1)
return 0;
}
public static void main(String[] args) {
int primeCheckVal = 0; // Value checked for prime
// Check primes for values 1 to 10
for (primeCheckVal = 1; primeCheckVal <= 10; ++primeCheckVal) {
if (isPrime(primeCheckVal, (primeCheckVal - 1)) == 1) {
System.out.println(primeCheckVal + " is prime.");
}
else {
System.out.println(primeCheckVal + " is not prime.");
}
}
}
}
My solution so far:到目前为止我的解决方案:
public class PrimeChecker {
// Returns 0 if value is not prime, 1 if value is prime
public static int isPrime(int testVal, int divVal) {
int resultVal = 0;
if ((testVal == 0) || (testVal == 1)){
resultVal = 0;
}// Base case 1: 0 and 1 are not prime, testVal is not prime
else if (divVal == 1) {
resultVal = 1;
}// Base case 2: testVal only divisible by 1, testVal is prime
else {
if((testVal % divVal != 0) && (testVal / divVal == 1)) {
isPrime(testVal, (divVal-1));
}
else {
resultVal = 1;
}
}
return resultVal;
}
public static void main(String[] args) {
int primeCheckVal = 0; // Value checked for prime
// Check primes for values 1 to 10
for (primeCheckVal = 1; primeCheckVal <= 10; ++primeCheckVal) {
if (isPrime(primeCheckVal, (primeCheckVal - 1)) == 1) {
System.out.println(primeCheckVal + " is prime.");
}
else {
System.out.println(primeCheckVal + " is not prime.");
}
}
}
}
Change the if/else block 更改if / else块
if((testVal % divVal != 0) && (testVal / divVal == 1)) {
isPrime(testVal, (divVal-1));
}
else {
resultVal = 1;
}
to 至
if (testVal % divVal != 0) {
return isPrime(testVal, (divVal-1));
} else {
resultVal = 0;
}
Basically, you've forgotten to return the result of your recursion, so the code carries on to return the wrong thing. 基本上,您忘记了返回递归的结果,因此代码继续执行以返回错误的内容。 If
testVal % divVal == 0
, the number is non-prime so you return zero. 如果
testVal % divVal == 0
,则该数字为非素数,因此您返回零。 Also, don't use ints that only take the value of zero or one; 另外,不要使用仅取零或一的整数; use a
boolean
. 使用
boolean
。
According your question, I think the following code would be easier.根据你的问题,我认为下面的代码会更容易。
public class PrimeChecker {
// Returns 0 if value is not prime, 1 if value is prime
public static int isPrime(int testVal, int divVal) {
// Base case 1: 0 and 1 are not prime, testVal is not prime
if (testVal <= 1) {
return 0;
}
// Base case 2: testVal only divisible by 1, testVal is prime
if (divVal == 1) {
return 1;
}
// Recursive Case
// Check if testVal can be evenly divided by divVal
// Hint: use the % operator
if (testVal % divVal == 0) {
return 0;
}
// If not, recursive call to isPrime with testVal and (divVal - 1)
return isPrime(testVal, divVal - 1);
}
public static void main(String[] args) {
int primeCheckVal; // Value checked for prime
// Check primes for values 1 to 10
for (primeCheckVal = 1; primeCheckVal <= 10; ++primeCheckVal) {
if (isPrime(primeCheckVal, (primeCheckVal - 1)) == 1) {
System.out.println(primeCheckVal + " is prime.");
}
else {
System.out.println(primeCheckVal + " is not prime.");
}
}
}
}
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