Java的：划分成若干部分等于或接近对方 [英] Java: divide number into equals parts or approximate to each other

问题描述

我有一个NUMER = 20。我要划分这个数字到N等于部分或接近对方。 Ñ​​也许从1至20个

I have a numer = 20. And I want to divide this number into N equals parts or approximate to each other. N maybe from 1 to 20.

在我写这篇code：

ArrayList<Integer> c = new ArrayList<>();
int num = 20;
int count = N; //where N some number (from 1 to 20)
int val = (int) Math.floor(num / count);

for (int i = 0; i < count; i++) {
    c.add(val);
}
c.set(0, c.get(0) + num - sum(c));

它工作正常，例如，当计数= N = 3。我得到08分，6，6（8 + 6 + 6 = 20）。但是，如果算上= N = 12我得到一个结果：9，1，1，1，1，1，1，1，1，1，1，1（总和= 20）。我想这个结果将在明年1，2，2，2，2，2，2，2，2，1，1，1（总和= 20）。这可能吗？

It works fine, for example, when count = N = 3. I get: 8, 6, 6 (8+6+6=20). But if count = N = 12 I get next result: 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 (sum = 20). I want that result will be next: 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1 (sum = 20). Is that possible?

推荐答案

您计算的大小的下一步的一部分，那么减少的值，并重复：

You calculate the size of the next part, then reduce the values and repeat:

private static int[] splitIntoParts(int whole, int parts) {
    int[] arr = new int[parts];
    int remain = whole;
    int partsLeft = parts;
    for (int i = 0; partsLeft > 0; i++) {
        int size = (remain + partsLeft - 1) / partsLeft; // rounded up, aka ceiling
        arr[i] = size;
        remain -= size;
        partsLeft--;
    }
    return arr;
}

如果你想，该方法可被挤压，但正如上面留下它是更好，因为它把这些参数作为不可改变的，明确的逻辑：

If you want, the method can be squeezed, though leaving it as above is better because it treats the parameters as immutable and clarifies the logic:

private static int[] splitIntoParts(int whole, int parts) {
    int[] arr = new int[parts];
    for (int i = 0; i < arr.length; i++)
        whole -= arr[i] = (whole + parts - i - 1) / (parts - i);
    return arr;
}

测试的

for (int parts = 0; parts <= 25; parts++)
    System.out.println(parts + ": " + Arrays.toString(splitIntoParts(20, parts)));

输出的

0: []
1: [20]
2: [10, 10]
3: [7, 7, 6]
4: [5, 5, 5, 5]
5: [4, 4, 4, 4, 4]
6: [4, 4, 3, 3, 3, 3]
7: [3, 3, 3, 3, 3, 3, 2]
8: [3, 3, 3, 3, 2, 2, 2, 2]
9: [3, 3, 2, 2, 2, 2, 2, 2, 2]
10: [2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
11: [2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1]
12: [2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1]
13: [2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1]
14: [2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1]
15: [2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
16: [2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
17: [2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
18: [2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
19: [2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
20: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
21: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0]
22: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0]
23: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0]
24: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0]
25: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0]

注意 0 返回空数组。添加如果语句，如果它失败了。负值将与 NegativeArraySizeException 失败。正如你所看到的，有太多的部分将 0的简单收益部分尺寸。再次，增加如果语句如果失败了

Notice that 0 returns empty array. Add if statement if it should fail. Negative values will fail with NegativeArraySizeException. As you can see, too many parts will simple return parts of 0 size. Again, add if statement if that should fail

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