问题
Thanks for the help from Zirak In my previous post i implemented the following in JavaScript:
var arr1 =[0,1,2,3];
var arr2 =["ac", "bc", "ad", "e"];
var result = arr1 .sort(function(i, j){return arr2[i].localeCompare(arr2[j])})
document.write(result );
The way to achieve this is quite compact in JavaScript, can a java implementation of this be also achieved by such simplicity? I could only think of implementing the Comparable interface like the following:
public class testCompare {
public static String[] arr2={"ac", "bc", "ad", "e"};
public static Obj[] arr1={new Obj(0), new Obj(1), new Obj(2), new Obj(3)};
static class Obj implements Comparable{
int index=0;
public Obj(int i){
index=i;
}
@Override
public int compareTo(Object o) {
return arr2[index].compareTo(arr2[((Obj)o).index]);
}
}
}
but if the array have X many items, then I will have to create X many Objs, is there another way that I could achieve this more simply? Another question is, if I do the above method what would be the time complexity for the sorting both in java and in JavaScript, are they all O(n^2)
? Thanks a lot
回答1:
public class MyComparator implements Comparator<Integer> {
@Override
public int compare(Integer i1, Integer i2) {
return arr2[i1.intValue()].compareTo(arr2[i2.intValue()]);
}
}
Arrays.sort(arr1, new MyComparator());
This is the equivalent of the JavaScript sort. The Comparator object is used as the callback function is used in JavaScript.
回答2:
Try using a TreeMap<String, Integer>
(assuming you want to sort integers) which means all entries are sorted by their string key:
SortedMap<String, Integer> map = new TreeMap<String, Integer>();
map.put("ac", 0);
map.put("bc", 1);
map.put("ad", 2);
map.put("e", 3);
for( Map.Entry<String, Integer> entry : map.entrySet() )
{
System.out.println(entry.getKey() + " - " + entry.getValue());
}
Output:
ac - 0
ad - 2
bc - 1
e - 3
To sort an array and get the new order of the previous indices you could iterate over the array and add the indices as Integer objects to the map:
String[] input = {"ab", "bc", "ad" , "e" };
SortedMap<String, Integer> map = new TreeMap<String, Integer>();
for( int i = 0; i < input.length; ++i )
{
map.put(input[i], i); //or use values from another array, e.g. map.put(inputKeys[i], inputValues[i]);
}
If you need to sort the keys by anything else but the natural order, you can add a Comparator<String>
to the TreeMap
constructor.
回答3:
public class SortA1byA2array
{
public static void main (String[] args)
{
int[] arr1={2,1,2,5,7,1,9,8,3,6,8,8};
int[] arr2={2,1,8,3};
TreeMap hm=new TreeMap();
int count=1;
for(int i=0;i<arr1.length;i++){
if(hm.containsKey(arr1[i])){
hm.put(arr1[i], ++count);
}
else{
count=1;
hm.put(arr1[i],count);
}
}
for(int i=0;i<arr2.length;i++){
if(hm.containsKey(arr2[i])){
for(int j=0;j<(Integer)hm.get(arr2[i]);j++){
System.out.println(arr2[i]);
}
hm.remove(arr2[i]);
}
}
Iterator it = hm.entrySet().iterator();
while (it.hasNext()) {
Map.Entry pairs = (Map.Entry)it.next();
System.out.println(pairs.getKey());
it.remove();
}
}
}
回答4:
In response to the second part of you question: Arrays.sort
in Java has guaranteed O(n log n) time complexity, as specified in the API.
来源:https://stackoverflow.com/questions/5898690/java-sort-array1-based-on-array2