Java 将封闭范围之间的奇数添加到数组
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Add odd numbers between a closed range to an array
提问by user6248190
I am doing code challenge where given two integers land r, I have to print all the odd numbers between iand r(iand rinclusive). The function must return an array of integers denoting the odd numbers between land r.
我做的代码的挑战，其中给出了两个整数l和r，我不得不之间打印所有的奇数i和r（i和r含）。该函数必须返回一个整数数组，表示l和之间的奇数r。
This is what I have so far
这是我到目前为止
static int[] oddNumbers(int l, int r) {
int[] theArray = new int[r];
for (int i = l; i < theArray.length; i++) {
if (i % 2 != 0) {
for (int n = 0; n < theArray.length; n++) {
theArray[n] = i;
}
}
}
return theArray;
}
So at the moment this code if you give it values 2 and 5 should return 3,5. However, this only returns 33333. What am I doing wrong? And how can I improve this code?
所以目前这个代码如果你给它值 2 和 5 应该返回 3,5。但是，这只返回 33333。我做错了什么？我该如何改进这段代码？
采纳答案by Tim Biegeleisen
I would approach this by advancing the lower bound in the range to the next odd number, if not already odd, and then proceeding from there.
我会通过将范围内的下限推进到下一个奇数来解决这个问题，如果不是奇数，然后从那里开始。
static int[] oddNumbers(int l, int r) {
if (r <= l) return null;
l = (l % 2) == 0 ? l + 1 : l;
int size = ((r - l) / 2) + 1;
int[] theArray = new int[size];
for (int i=0; i < size; ++i) {
theArray[i] = l + (i*2);
}
return theArray;
}
The real difficulty here is in formulating the logic to map a range of odd numbers onto a flat array. Once we have done this, we can see that populating the array only requires a simple for loop.
这里真正的困难在于制定将一系列奇数映射到平面数组的逻辑。完成此操作后，我们可以看到填充数组只需要一个简单的 for 循环。
回答by oleg.cherednik
static int[] oddNumbers(int l, int r) {
int[] theArray = new int[((r - l) / 2) + 1];
for (int i = l % 2 == 0 ? l : l + 1, j = 0; i <= r; i += 2, j++)
theArray[j] = i;
return theArray;
}
回答by Andrew Tobilko
I would delegate the work to a stream or list which can easily and dynamically manage the size of a selection.
我会将工作委托给一个流或列表，它们可以轻松动态地管理选择的大小。
Why should we bother about the size of a resulting array if we could make a simple transformation from an appropriate structure at the end?
如果我们可以在最后从一个适当的结构进行简单的转换，为什么我们还要为结果数组的大小而烦恼呢？
static int[] oddNumbers(int l, int r) {
return IntStream.iterate(l, i -> i <= r, i -> ++i)
.filter(i -> i % 2 != 0)
.toArray();
}
回答by roxch
We are in 2018 and do already have the amazing Streamssince Java 8! With that you can solve such things in a single line! So it's worth it to take a look at it:
我们在 2018 年，并且已经拥有Streams自 Java 8 以来的惊人！有了它，您可以在一行中解决此类问题！所以值得一看：
static int[] oddNumbers(int i, int j){
return IntStream.rangeClosed(i, j).filter(num -> num % 2 == 1).toArray();
}
(unless you really want to learn algorithms and not just solve the challenge!)
（除非你真的想学习算法而不仅仅是解决挑战！）
