英文:
Shuffle list within n size of bins in list
问题
我有一个Java列表，示例：
a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]
我想要对它进行洗牌，但是要在n大小的桶内进行。所谓的桶，我是指我想要先洗牌前5个元素，然后是接下来的5个元素，依此类推...
因此，其中一种期望的结果将是：
sorted = [3,5,4,2,1, 8,9,7,10,6, 14,11,12,15,13, 19,20,17,18,16, 22,21]
如何高效地实现这个目标？
英文:
I have a java list, example:
a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]
I want to shuffle it but within n size bins within. From bin, I meant, I want to shuffle first 5 elements, then next 5 and so on..
So, one of the the expected result will be:
sorted = [3,5,4,2,1, 8,9,7,10,6, 14,11,12,15,13, 19,20,17,18,16, 22,21]
How can I do it efficiently
答案1
得分: 4
将数组放入列表中并对子列表进行洗牌：
Integer[] array = new Integer[] {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22
};
List<Integer> list = Arrays.asList(array);
int binSize = 5;
for (int i = 0, n = list.size(); i < n; i += binSize) {
int j = Math.min(i + binSize, n);
Collections.shuffle(list.subList(i, j));
}
array = list.toArray(new Integer[] {});
System.out.println(Arrays.toString(array));
[2, 1, 3, 4, 5, 6, 8, 10, 7, 9, 13, 11, 15, 14, 12, 19, 16, 17, 20, 18, 22, 21]
英文:
Put the array into a list and shuffle the sublists:
Integer[] array = new Integer [] {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22
};
List<Integer> list = Arrays.asList(array);
int binSize = 5;
for (int i = 0, n = list.size(); i < n; i += binSize) {
int j = Math.min(i + binSize, n);
Collections.shuffle(list.subList(i, j));
}
array = list.toArray(new Integer[] { });
System.out.println(Arrays.toString(array));
[2, 1, 3, 4, 5, 6, 8, 10, 7, 9, 13, 11, 15, 14, 12, 19, 16, 17, 20, 18, 22, 21]
答案2
得分: 0
Just shuffle each subList defined by the start and end of the bins. Since the subList provides a view of the list, they are shuffled in place so there is no need to create a new collection.
List<Integer> vals = new ArrayList<>(
List.of(1,2,3,4,5,10,20,30,40,50,100,200,300,400,500,1000,2000));
shuffleBins(vals);
System.out.println(vals);
Prints
[4, 5, 2, 3, 1, 10, 40, 20, 50, 30, 400, 200, 300, 100, 500, 2000, 1000]
The method
public static void shuffleBins(int binSize,
List<Integer> items) {
int start = 0;
for (int i = 0; i < items.size()/5; i++) {
Collections.shuffle(items.subList(start, start+binSize ));
start += binSize;
}
// finish up when items.size() % binSize != 0
Collections.shuffle(items.subList(start, items.size() ));
}
If you want to make it more versatile you can provide a set of indices indicating the bin starts.
public static void shuffleBins(List<Integer> binStarts,
List<Integer> items) {
int start = 0;
for (int i = 0; i < binStarts.size(); i++) {
Collections
.shuffle(items.subList(start, binStarts.get(i)));
start = binStarts.get(i) + 1;
}
Collections.shuffle(items.subList(start, items.size()));
}
英文:
Just shuffle each subList defined by the start and end of the bins. Since the subList provides a view of the list, they are shuffled in place so there is no need to create a new collection.
List<Integer> vals = new ArrayList<>
(List.of(1,2,3,4,5,10,20,30,40,50,100,200,300,400,500,1000,2000));
shuffleBins(vals);
System.out.println(vals);
Prints
[4, 5, 2, 3, 1, 10, 40, 20, 50, 30, 400, 200, 300, 100, 500, 2000, 1000]
The method
public static void shuffleBins(int binSize,
List<Integer> items) {
int start = 0;
for (int i = 0; i < items.size()/5; i++) {
Collections.shuffle(items.subList(start, start+binSize ));
start += binSize;
}
// finish up when items.size() % binSize != 0
Collections.shuffle(items.subList(start, items.size() ));
}
If you want to make it more versatile you can provide a set of indices indicating the bin starts.
public static void shuffleBins(List<Integer> binStarts,
List<Integer> items) {
int start = 0;
for (int i = 0; i < binStarts.size(); i++) {
Collections
.shuffle(items.subList(start, binStarts.get(i)));
start = binStarts.get(i) + 1;
}
Collections.shuffle(items.subList(start, items.size()));
}
</details>
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