我在编写下面的代码时遇到了问题：
编写一个名为removeDuplicates的静态方法，该方法将整数数组作为输入，并返回一个新的整数数组，其中所有重复项都已删除。
例如，如果输入数组具有元素{4、3、3、4、5、2、4}，则结果数组应为{4、3、5、2}
到目前为止，我的代码如下：
public static int[] removeDuplicates(int []s){
int [] k = new int[s.length];
k[0]=s[0];
int m =1;
for(int i=1;i<s.length;++i){
if(s[i]!=s[i-1]){
k[m]=s[i];
++m;
}//endIF
}//endFori
return k;
}//endMethod
public void removeDupInIntArray(int[] ints){
Set<Integer> setString = new LinkedHashSet<Integer>();
for(int i=0;i<ints.length;i++){
setString.add(ints[i]);
}
System.out.println(setString);
}
试试这个 -
public static int[] removeDuplicates(int []s){
int result[] = new int[s.length], j=0;
for (int i : s) {
if(!isExists(result, i))
result[j++] = i;
}
return result;
}
private static boolean isExists(int[] array, int value){
for (int i : array) {
if(i==value)
return true;
}
return false;
}
首先，您应该知道不包括重复项（dups）的长度：初始长度减去重复项的数量。 然后创建新的具有正确长度的数组。 然后检查list[]中的每个元素是否存在重复项，如果找到重复项-检查下一个元素，如果没有找到重复项-将元素复制到新数组中。
public static int[] eliminateDuplicates(int[] list) {
int newLength = list.length;
// find length w/o duplicates:
for (int i = 1; i < list.length; i++) {
for (int j = 0; j < i; j++) {
if (list[i] == list[j]) { // if duplicate founded then decrease length by 1
newLength--;
break;
}
}
}
int[] newArray = new int[newLength]; // create new array with new length
newArray[0] = list[0]; // 1st element goes to new array
int inx = 1; // index for 2nd element of new array
boolean isDuplicate;
for (int i = 1; i < list.length; i++) {
isDuplicate = false;
for (int j = 0; j < i; j++) {
if (list[i] == list[j]) { // if duplicate founded then change boolean variable and break
isDuplicate = true;
break;
}
}
if (!isDuplicate) { // if it's not duplicate then put it to new array
newArray[inx] = list[i];
inx++;
}
}
return newArray;
}
int numbers[] = {1,2,3,4,1,2,3,4,5,1,2,3,4};
numbers = java.util.stream.IntStream.of(numbers).distinct().toArray();
import static ch.lambdaj.Lambda.selectDistinct;
import java.util.Arrays;
import java.util.List;
public class DistinctList {
public static void main(String[] args) {
List<Integer> numbers = Arrays.asList(1,3,4,2,1,5,6,8,8,3,4,5,13);
System.out.println("List with duplicates: " + numbers);
System.out.println("List without duplicates: " + selectDistinct(numbers));
}
}
List with duplicates: [1, 3, 4, 2, 1, 5, 6, 8, 8, 3, 4, 5, 13]
List without duplicates: [1, 2, 3, 4, 5, 6, 8, 13]
一行代码即可得到一个不同的列表，这只是一个简单的例子，但使用该库，您可以解决更多问题。
selectDistinct(numbers)
public int[] removeRepetativeInteger(int[] list){
if(list.length == 0){
return null;
}
if(list.length == 1){
return list;
}
ArrayList<Integer> numbers = new ArrayList<>();
for(int i = 0; i< list.length; i++){
if (!numbers.contains(list[i])){
numbers.add(list[i]);
}
}
Iterator<Integer> valueIterator = numbers.iterator();
int[] resultArray = new int[numbers.size()];
int i = 0;
while (valueIterator.hasNext()) {
resultArray[i] = valueIterator.next();
i++;
}
return resultArray;
}
public class Solution4 {
public static void main(String[] args) {
int[] a = {1,1,2,3,4,5,6,6,7,8};
int countwithoutDuplicates = lengthofarraywithoutDuplicates(a);
for(int i = 0 ; i < countwithoutDuplicates ; i++) {
System.out.println(a[i] + " ");
}
}
private static int lengthofarraywithoutDuplicates(int[] a) {
int countwithoutDuplicates = 1 ;
for (int i = 1; i < a.length; i++) {
if( a[i] != a[i-1] ) {
a[countwithoutDuplicates++] = a[i];
}//if
}//for
System.out.println("length of array withpout duplicates = >" + countwithoutDuplicates);
return countwithoutDuplicates;
}//lengthofarraywithoutDuplicates
}
在Python中：
def lengthwithoutduplicates(nums):
if not nums: return 0
if len(nums) == 1:return 1
# moving backwards from last element i.e.len(a) -1 to first element 0 and step is -1
for i in range(len(nums)-1,0,-1):
# delete the repeated element
if nums[i] == nums[i-1]: del nums[i]
# store the new length of the array without the duplicates in a variable
# and return the variable
l = len(a)
return l
a = [1, 1, 2, 3, 4, 5, 6, 6, 7, 8];
l = lengthwithoutduplicates(a)
for i in range(1,l):print(i)
a = [1, 1, 2, 3, 4, 5, 6, 6, 7, 8]
aa = [ ch for i, ch in enumerate(a) if ch not in a[:i] ]
print(aa) # output => [1, 2, 3, 4, 5, 6, 7, 8]
public class DuplicatesFromArray {
public static void main(String args[]) {
int[] withDuplicates = { 1, 2, 3, 1, 2, 3, 4, 5, 3, 6 };
// complexity of this solution is O[n]
duplicates(withDuplicates);
}
//Complexity of this method is O(n)
public static void duplicates(int[] input) {
HashSet<Integer> nums = new HashSet<Integer>();
List<Integer> results = new ArrayList<Integer>();
List<Integer> orderedFiltered = new ArrayList<Integer>();
for (int in : input) {
if (nums.add(in) == false) {
results.add(in);
} else {
orderedFiltered.add(in);
}
}
out.println(
"Ordered and filtered elements found in the array are : " + Arrays.toString(orderedFiltered.toArray()));
out.println("Duplicate elements found in the array are : " + Arrays.toString(results.toArray()));
}
/**
* Generic method to find duplicates in array. Complexity of this method is O(n)
* because we are using HashSet data structure.
*
* @param array
* @return
*/
public static <T extends Comparable<T>> void getDuplicates(T[] array) {
Set<T> dupes = new HashSet<T>();
for (T i : array) {
if (!dupes.add(i)) {
System.out.println("Duplicate element in array is : " + i);
}
}
}
}
public static void deleteDups(int a []) {
HashSet<Integer> numbers = new HashSet<Integer>();
for(int n : a)
{
numbers.add(n);
}
for(int k : numbers)
{
System.out.println(k);
}
System.out.println(numbers);
}
public static void main(String[] args) {
int a[]={2,3,3,4,4,5,6};
RemoveDuplicate.deleteDups(a);
}
}
o/p is 2
3
4
5
6
[2, 3, 4, 5, 6]
你也可以将数组元素放入一个Set中，其语义精确地表示它不包含重复元素。
Set set = new HashSet( Arrays.asList( s ) ) 这段代码无法编译。 - Subhrajyoti MajumderArrays.asList()将int[]转换为Integer[]。 - assylias