英文:
How to find the sum of divisors using Java streams?
问题
我正在尝试将这个函数转换为使用Java 8的新语法。希望能减少代码行数，或许能使代码更清晰。
public int divisorSum(int n) {
int sum = 0;
for (int i = 1; i <= n; i++) {
if (n % i == 0) {
sum = Integer.sum(sum, i);
}
}
return sum;
}
我尝试了以下内容：
IntStream.range(0, n).forEach(i -> ... )
但根据这个帖子中Tezra的评论，显然不建议使用lambda进行循环。
英文:
I am trying to convert this function to use Java 8 new syntax. Hopefully it will reduce the number of lines and perhaps make things clearer.
public int divisorSum(int n) {
int sum = 0;
for(int i = 1; i <= n; i ++) {
if(n % i == 0) {
sum = Integer.sum(sum, i);
}
}
return sum;
}
I tried this:
IntStream.range(0, n).forEach(i -> ... )
But according to a comment on this post by Tezra apparently it is not advisable to loop using lambdas.
答案1
得分: 4
这是Java 8流的实现：
public int divisorSum(int n) {
return IntStream.rangeClosed(1, n).filter(i -> n % i == 0).sum();
}
注意，rangeClosed与你的示例一样，包括 n。range() 则排除第二个参数（它只会包括到 n-1）。
英文:
Here's the Java 8 streams implementation:
public int divisorSum(int n) {
return IntStream.rangeClosed(1, n).filter(i -> n % i == 0).sum();
}
Note that rangeClosed, like your example, includes n. range() excludes the second parameter (it would only include up to n-1).
答案2
得分: 3
你可以使用IntStream、filter和IntStream::sum来实现相同的结果，这个流直接返回int，因为这个流是拆箱的：
int sum = IntStream.rangeClosed(1, n).filter(i -> n % i == 0).sum();
英文:
You can achieve the same result using IntStream, filter and IntStream::sum that directly returns int since this stream is unboxed:
int sum = IntStream.rangeClosed(1, n).filter(i -> n % i == 0).sum();
答案3
得分: 3
你可以像这样做：
public static int divisorSum(int n) {
return IntStream.rangeClosed(1, n)
.filter(i -> n % i == 0)
.sum();
}
英文:
You can do something like this
public static int divisorSum(int n) {
return IntStream.rangeClosed(1, n)
.filter(i -> n % i == 0)
.sum();
}
答案4
得分: 1
这可能会有所帮助。
int sum1 = java.util.stream.IntStream.range(1, n + 1).filter(x -> n % x == 0).reduce(0, (x, y) -> x + y);
或者
int sum1 = java.util.stream.IntStream.range(1, n + 1).filter(x -> n % x == 0).sum();
英文:
This can be helpful.
int sum1 = java.util.stream.IntStream.range(1, n + 1).filter(x -> n % x == 0).reduce(0, (x, y) -> x + y);
or
int sum1 = java.util.stream.IntStream.range(1, n + 1).filter(x -> n % x == 0).sum();
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