java 如何将字符串数组转换为唯一值数组？
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How do I convert an array of strings to a array of unique values?
提问by djangofan
In Java, how do I convert an array of strings to a array of unique values?
在 Java 中，如何将字符串数组转换为唯一值数组？
If I have this array of Strings:
如果我有这个字符串数组：
String[] test = {"1","1","1","2"}
And I want to end up with:
我想结束：
String[] uq = {"1","2"}
采纳答案by Taig
If you're going with the HashSet-approach (which seems pretty handy) you should use a LinkedHashSetinstead of a HashSetif you want to maintain the array's order!
如果您要使用HashSet-approach（这看起来很方便），您应该使用 aLinkedHashSet而不是 aHashSet来维护数组的顺序！
Set<String> temp = new LinkedHashSet<String>( Arrays.asList( array ) );
String[] result = temp.toArray( new String[temp.size()] );
回答by ChssPly76
Quick but somewhat inefficient way would be:
快速但有点低效的方法是：
Set<String> temp = new HashSet<String>(Arrays.asList(test));
String[] uq = temp.toArray(new String[temp.size()]);
回答by djangofan
I tried all the answers on this page and none worked as-is. So, here is how I solved it, inspired by the answers from Taigand akuhn:
我尝试了此页面上的所有答案，但没有一个按原样工作。因此，受Taig和akuhn的回答启发，我解决了这个问题：
import groovy.io.*;
def arr = ["5", "5", "7", "6", "7", "8", "0"]
List<String> uniqueList = new ArrayList<String>(
new LinkedHashSet<String>( arr.asList() ).sort() );
System.out.println( uniqueList )
回答by maximdim
String[] test = {"1","1","1","2"};
java.util.Set result = new java.util.HashSet(java.util.Arrays.asList(test));
System.out.println(result);
回答by Stephen C
An alternative to the HashSet approach would be to:
HashSet 方法的替代方法是：
Sort the input array
Count the number of non-duplicate values in the sorted array
Allocate the output array
Iterate over the sorted array, copying the non-duplicate values to it.
对输入数组进行排序
计算排序数组中非重复值的数量
分配输出数组
迭代已排序的数组，将非重复值复制到其中。
The HashSet approach is O(N)on average assuming that 1) you preallocate the HashSet with the right size and 2) the (non-duplicate) values in the input array hash roughly evenly. (But if the value hashing is pathological, the worst case is O(N**2)!)
HashSet 方法O(N)平均假设 1) 您预先分配了具有正确大小的 HashSet 和 2) 输入数组中的（非重复）值大致均匀地散列。（但如果值散列是病态的，最坏的情况是O(N**2)！）
The sorting approach is O(NlogN)on average.
排序方法是O(NlogN)平均的。
The HashSet approach takes more memory on average.
HashSet 方法平均需要更多内存。
If you are doing this infrequently ORfor really large "well behaved" input arrays, the HashSet approach is probably better. Otherwise, it could be a toss-up which approach is better.
如果您很少这样做，或者对于非常大的“表现良好”的输入数组，则 HashSet 方法可能更好。否则，这可能是一种折腾，哪种方法更好。
回答by Anon.
An easy way is to create a set, add each element in the array to it, and then convert the set to an array.
一种简单的方法是创建一个集合，将数组中的每个元素添加到其中，然后将集合转换为数组。
回答by victor hugo
List list = Arrays.asList(test);
Set set = new HashSet(list);
String[] uq = set.toArray();
回答by ThibaudL
Just found a nicer way in Java 8 :
刚刚在 Java 8 中找到了一个更好的方法：
Arrays.stream(aList).distinct().toArray(String[]::new)
回答by Jianfeng Tian
here is my solution:
这是我的解决方案：
int[] A = {2, 1, 2, 0, 1};
Arrays.sort(A);
ArrayList<Integer> B = new ArrayList<Integer>();
for (int i = 0; i < A.length; i++) {
if (i == A.length-1) {
B.add(A[i]);
}
else if (A[i] != A[i+1]) {
B.add(A[i]);
}
}
回答by Hari
String[] getDistinctElementsArray(String[] arr){
StringBuilder distStrings = new StringBuilder();
distStrings.append(arr[0] + " ");
for(int i=1;i<arr.length;i++){
if( arr[i].equals(arr[i-1])){}
else{
distStrings.append(arr[i] + " ");
}
}
return distStrings.toString().split(" ");
}
