我有一个像这样的数组元素：
int arr[] = {1,1,2,2,3,3,4,4};
我想从中删除重复的元素。在互联网上搜索并了解了ArrayUtil类。你可以通过告诉它的用法来帮助我 - 我怎样才能在输出中得到这样的数组：
arr[] = {1,2,3,4};
答案 0 :(得分：5)
这是Element Distinctness Problem的下一步，在此主题中详细讨论：Find duplicates in an array，包括问题的下限（如果没有涉及哈希集，则不能比O(nlogn)做得更好）。
如果您不愿意使用哈希集来检查您已经看过的所有元素，那么最好的办法是对数组进行排序，然后迭代它 - 所有重复的元素将彼此相邻。
public static int[] justUniques(int[] arr) {
if (arr == null || arr.length == 0) return arr;
Arrays.sort(arr);
int n = 1;
for (int i = 1; i < arr.length; i++) {
if (arr[i] != arr[i-1]) n++;
}
int[] res = new int[n];
res[0] = arr[0];
n = 1;
for (int i = 1; i < arr.length; i++) {
if (arr[i] != arr[i-1]) res[n++] = arr[i];
}
return res;
}
请注意，上述的简单变体也可以就地执行，而无需创建新数组。
此解决方案为O(nlogn)，因此是最佳解决方案。如果您不愿意使用Arrays.sort()，您可以实施自己的排序算法（这非常简单）。
另一个相关主题提出类似问题并附加限制：Removing the duplicates from an array without disturbing the order of elements without using Sets
答案 1 :(得分：2)
为这个问题找到了一个非常好的解决方案：它完美无缺。对于那些仍在寻找这个问题的答案的人，你可以使用下面这段代码。
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class DuplicatesRemove {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter the array size :");
int size = Integer.parseInt(br.readLine());
int[] arr = new int[size];
// Creating the array
for (int i = 0; i < size; i++) {
System.out.print("Enter the element arr[" + i + "] = ");
arr[i] = Integer.parseInt(br.readLine());
System.out.println();
}
// displaying the array - it may contain elements in unsorted manner
System.out.println("Before Sorting :");
for (int i = 0; i < size; i++) {
System.out.println("Element arr[" + i + "] = " + arr[i]);
}
System.out
.println("*****************************************************");
// Logic for sorting the elements in the array
for (int i = 0; i < size; i++) {
for (int j = 1; j < size - i; j++) {
if (arr[j - 1] > arr[j]) {
int temp = arr[j];
arr[j] = arr[j - 1];
arr[j - 1] = temp;
}
}
}
// Printing the sorted elements - it may contain duplicate elements
System.out.println("After Sorting :");
for (int i = 0; i < size; i++) {
System.out.println("Element arr[" + i + "] = " + arr[i]);
}
System.out
.println("*****************************************************");
// Logic for removing the duplicate elements
int compare = 0;
arr[compare] = arr[0];
for (int i = 1; i < size; i++) {
if (arr[compare] != arr[i]) {
compare++;
arr[compare] = arr[i];
}
}
System.out.println("Array After removing duplicate elements is :");
for (int i = 0; i <= compare; i++) {
System.out.println("Element arr[" + i + "] = " + arr[i]);
}
}
}
答案 2 :(得分：0)
public int[] removeDuplicates(int[] arr) {
int[] res = new int[arr.length];
int index = 0;
for (int num : arr) {
if (res.indexOf(num) == -1)
res[index++] = num;
}
return res;
}
这是一个次优解决方案，但它不需要对数组进行任何排序。我创建了一个新数组，遍历原始数组中的项目，如果它们已经存在，则将项目添加到新数组中。
答案 3 :(得分：0)
public static int[] removeDuplicates(int[] input){
int j = 0;
int i = 1;
//return if the array length is less than 2
if(input.length < 2){
return input;
}
while(i < input.length){
if(input[i] == input[j]){
i++;
}else{
input[++j] = input[i++];
}
}
int[] output = new int[j+1];
for(int k=0; k<output.length; k++){
output[k] = input[k];
}
return output;
}
来源：http://java2novice.com/java-interview-programs/remove-duplicates-sorted-array/
答案 4 :(得分：0)
package practice1;
import java.util.Scanner;
public class RemoveArrayDuplicatesElements {
public static void main(String[] args) {
int i, j, k, size,same = 0;
System.out.println("\nEnter array size : ");
@SuppressWarnings("resource")
Scanner sc = new Scanner(System.in);
size = sc.nextInt();
int[] arr = new int[size+1];
System.out.println("\nAccept Numbers : ");
for (i = 0; i < size; i++)
arr[i] = sc.nextInt();
System.out.println("\nArray with Unique list : ");
for (i = 0; i < size; i++) {
for (j = i + 1; j < size;) {
if (arr[j] == arr[i]) {
same++;
for (k = j; k < size; k++) {
arr[k] = arr[k + 1];
}
size--;
} else
j++;
}
}
for (int g = 0; g < arr.length; g++) {
System.out.println(arr[g]);
}
}
}
答案 5 :(得分：0)
int flag = 0, k = 0;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arrAns.length; j++) {
if (arr[i] == arrAns[j]) {
flag = 0;
break;
}
flag=1;
}
if (flag == 1) {
arrAns[k] = arr[i];
k++;
}
flag = 0;
}
答案 6 :(得分：0)
public class UniqueElementinAnArray
{
public static void main(String[] args)
{
int[] a = {10,10,10,10,10,100};
int[] output = new int[a.length];
int count = 0;
int num = 0;
//Iterate over an array
for(int i=0; i<a.length; i++)
{
num=a[i];
boolean flag = check(output,num);
if(flag==false)
{
output[count]=num;
++count;
}
}
//print the all the elements from an array except zero's (0)
for (int i : output)
{
if(i!=0 )
System.out.print(i+" ");
}
}
/***
* If a next number from an array is already exists in unique array then return true else false
* @param arr Unique number array. Initially this array is an empty.
* @param num Number to be search in unique array. Whether it is duplicate or unique.
* @return true: If a number is already exists in an array else false
*/
public static boolean check(int[] arr, int num)
{
boolean flag = false;
for(int i=0;i<arr.length; i++)
{
if(arr[i]==num)
{
flag = true;
break;
}
}
return flag;
}
}
答案 7 :(得分：0)
使用以下方法：
/*
* Method to remove duplicates from array in Java, without using
* Collection classes e.g. Set or ArrayList. Algorithm for this
* method is simple, it first sort the array and then compare adjacent
* objects, leaving out duplicates, which is already in the result.
*/
public static int[] removeDuplicates(int[] numbersWithDuplicates) {
// Sorting array to bring duplicates together
Arrays.sort(numbersWithDuplicates);
int[] result = new int[numbersWithDuplicates.length];
int previous = numbersWithDuplicates[0];
result[0] = previous;
for (int i = 1; i < numbersWithDuplicates.length; i++) {
int ch = numbersWithDuplicates[i];
if (previous != ch) {
result[i] = ch;
}
previous = ch;
}
return result;
}
答案 8 :(得分：0)
这是删除重复元素而不应用排序和集合的最佳解决方案。
public static int[] removeElm(int arr[]) {
int[] tempArr = new int[arr.length];
int j = 0;
tempArr[j] = arr[0];
for (int i = 1; i < arr.length; i++) {
boolean check = false;
for (int k = 0; k < j + 1; k++) {
if (tempArr[k] != arr[i]) {
check = true;
} else {
check = false;
break;
}
}
if (check) {
tempArr[++j] = arr[i];
}
}
return tempArr;
}
答案 9 :(得分：-2)
package com.array;
import java.util.*;
class RemoveDuplicatesInArray{
public static void main(String[] args) {
Integer[] array = new Integer[10];
array[0] = 1;
array[1] = 2;
array[2] = 3;
array[3] = 3;
array[4] = 3;
array[5] = 3;
array[6] = 7;
array[7] = 7;
array[8] = 9;
array[9] = 9;
removeDuplicatesFromArray(array);
}
private static void removeDuplicatesFromArray(Integer[] array){
StringBuffer stringBuffer = new StringBuffer();
String arrayString = Arrays.toString(array);
for(int index =0 ; index <= arrayString.length(); index++){
try{
int number = Integer.parseInt(arrayString.charAt(index)+"");
if(!stringBuffer.toString().contains(number+"")){
if(stringBuffer.length()!=0)
stringBuffer.append(",");
stringBuffer.append(number);
}
}catch(Exception e){
}
}
String[] stringArray = stringBuffer.toString().split(",");
array = new Integer[stringArray.length];
for(int index = 0 ; index < stringArray.length ; index++){
array[index] = Integer.parseInt(stringArray[index]);
}
System.out.println(Arrays.toString(array));
}
}