074*_*ude 20 java arrays sorting
我试图按字典顺序对三个数组进行排序.阵列通过公共阵列彼此相关.如果我证明,它更容易解释:
int[] record = new int[4];
String [] colors = {"blue", "yellow", "red", "black"};
String [] clothes = {"shoes", "pants", "boots", "coat"};
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在控制台上打印时,我希望将它们放在类似下面的三列中:
Record Color Clothes
0 blue shoes
1 yellow pants
2 red boots
3 black coat
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Record Color Clothes
3 black coat
0 blue shoes
2 red boots
1 yellow pants
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Record Color Clothes
2 red boots
3 black coat
1 yellow pants
0 blue shoes
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我发现了一个类似于我的场景的先前答案,但是它比较了整数而不是字符串,而且我在使用该compareTo()方法时遇到了麻烦并且Arrays.sort()达到了我想要的输出.
任何帮助,将不胜感激!
在某些情况下,创建一个新类只是为了进行排序没有多大意义.
这里,是一个函数,可用于List<?>根据键列表(List<T implements Comparable>)对任意数量的任意类型列表()进行排序. 这里是Ideone示例.
下面是一个如何使用该函数对任意类型的多个列表进行排序的示例:
List<Integer> ids = Arrays.asList(0, 1, 2, 3);
List<String> colors = Arrays.asList("blue", "yellow", "red", "black");
List<String> clothes = Arrays.asList("shoes", "pants", "boots", "coat");
// Sort By ID
concurrentSort(ids, ids, colors, clothes);
// Sort By Color
concurrentSort(colors, ids, colors, clothes);
// Sort By Clothes
concurrentSort(clothes, ids, colors, clothes);
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输出:
// Sorted By ID:
ID: [0, 1, 2, 3]
Colors: [blue, yellow, red, black]
Clothes: [shoes, pants, boots, coat]
// Sorted By Color:
ID: [3, 0, 2, 1]
Colors: [black, blue, red, yellow]
Clothes: [coat, shoes, boots, pants]
// Sorted By Clothes:
ID: [2, 3, 1, 0]
Colors: [red, black, yellow, blue]
Clothes: [boots, coat, pants, shoes]
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public static <T extends Comparable<T>> void concurrentSort(
final List<T> key, List<?>... lists){
// Create a List of indices
List<Integer> indices = new ArrayList<Integer>();
for(int i = 0; i < key.size(); i++)
indices.add(i);
// Sort the indices list based on the key
Collections.sort(indices, new Comparator<Integer>(){
@Override public int compare(Integer i, Integer j) {
return key.get(i).compareTo(key.get(j));
}
});
// Create a mapping that allows sorting of the List by N swaps.
// Only swaps can be used since we do not know the type of the lists
Map<Integer,Integer> swapMap = new HashMap<Integer, Integer>(indices.size());
List<Integer> swapFrom = new ArrayList<Integer>(indices.size()),
swapTo = new ArrayList<Integer>(indices.size());
for(int i = 0; i < key.size(); i++){
int k = indices.get(i);
while(i != k && swapMap.containsKey(k))
k = swapMap.get(k);
swapFrom.add(i);
swapTo.add(k);
swapMap.put(i, k);
}
// use the swap order to sort each list by swapping elements
for(List<?> list : lists)
for(int i = 0; i < list.size(); i++)
Collections.swap(list, swapFrom.get(i), swapTo.get(i));
}
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注:在运行时间O(mlog(m) + mN),其中m是列表的长度,N是列表的数目.通常m >> N所以运行时间并不比仅对键进行排序更重要O(mlog(m)).
因为Record,Color并且Clothes似乎属于一起,我建议将它们一起移动到自定义对象中,例如
public class ClothesItem {
int record;
String color;
String clothes;
}
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然后你可以做出不同的Comparators来做不同的排序变体.
如果您需要使用多个数组保留当前结构,@ Jackson在这里有一个排序解决方案,它可以获得一系列已排序的索引,这样可以轻松获得您想要的结果.