找到数组的第一个非重复整数

Question

This program is giving output for all non-repeated elements but I need first one non-repeated element. I tried to keep if(flag==1) to break loop after the end of j loop, then I tested but it is not working for all cases

import java.util.Scanner;
public class first
{
    public static void main(String[] args) 
    {
        int n, flag = 0;
        Scanner s = new Scanner(System.in);
        System.out.print("Enter no. of elements you want in array:");
        n = s.nextInt();
        int a[] = new int[n];
        System.out.println("Enter all the elements:");
        for(int i = 0; i < n; i++)
        {
            a[i] = s.nextInt();
        }
        System.out.print("Non repeated first element is :");
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(i != j)
                {
                    if(a[i]!= a[j])
                    {
                        flag = 1;

                    }
                    else
                    {
                        flag = 0;
                        break;
                    }
                    if(flag == 1)
                    {
                     System.out.print(" ");   
                     System.out.println(a[i]);
                     break; 
                    }
                }
            }

        }

    }
}

Answer 1

You can construct two sets, singleSet and repeatedSet , respectively for elements appeared once and more than once. They can be created by doing one iteration on elements. Then, you do an a second iteration, to query which is the first element non-repeated:

int[] elements = { 1, 1, 2, 3, 3, 4 };
Set<Integer> singleSet = new HashSet<>();
Set<Integer> repeatedSet = new HashSet<>();

for (int e : elements) {
    if (repeatedSet.contains(e)) {
        continue;
    }
    if (singleSet.contains(e)) {
        singleSet.remove(e);
        repeatedSet.add(e);
    } else {
        singleSet.add(e);
    }
}

for (int e : elements) {
    if (singleSet.contains(e)) {
        return e;
    }
}

This solution is a O(n) solution, it should be faster than the nested-loop, which is O(n^2) .

---

You can also replace the singeSet by a singleList , and at the end, return the first element in the singleList , which avoid the 2nd iteration on elements. Thus the solution is even faster.

Answer 2

Following up on the idea of the two sets from @Mincong , I am adding here the solution he mentioned as faster.

int[] array = { 1, 1, 2, 3, 3, 4 };

Set<Integer> allValues = new HashSet<>(array.length);
Set<Integer> uniqueValues = new LinkedHashSet<>(array.length);

for (int value : array) {
    if (allValues.add(value)) {
        uniqueValues.add(value);
    }
    else {
        uniqueValues.remove(value);
    }
}

if (!uniqueValues.isEmpty()) {
    return uniqueValues.iterator().next();
}

Answer 3

First non-repeating integers in an array in Python

def non_repeating(arr):
    non_repeating = []
    for n in arr:
        if n in non_repeating:
            non_repeating.pop(non_repeating.index(n))
        else:
            non_repeating.append(n)

    return non_repeating[0] if non_repeating else None


print(non_repeating([1, 1, 1, 5, 2, 1, 3, 4, 2]))

Answer 4

def solution(self, list):
    count_map = {}

    for item in list:
         count_map[item] = count_map.get(item, 0) + 1

    for item in list:
        if count_map[item] == 1:
            return item

    return None

Answer 5

Here is a python code trying to achieve the same -

def singleNumber(nums: List[int]) -> int:
        from collections import defaultdict
        memory = defaultdict(int)
        for num in nums:
            memory[num] += 1
        for k,v in memory.items():
            if v == 1:
                return k

Answer 6

Javascript

function nonRepeatinInteger(arr) {
    let val = [], count = [];
    arr.forEach((item, pos) => {
        if (!val.includes(item)) {
            val.push(item);
            count[val.indexOf(item)] = 1;
        } else {
            count[val.indexOf(item)]++;
        }
    });
    return val[count.indexOf(Math.min(...count))];
}
console.log(nonRepeat([-1, 2, -1, 3, 2]));
console.log(nonRepeat([9, 4, 9, 6, 7, 4]));

Answer 7

 private int getFirstNonRepeating(int[] arr) {
    Set<Integer> set = new HashSet<>();
    ArrayList<Integer> list = new ArrayList<>();
    int min = 0;
    for (int i = 0; i <arr.length; i++) {
        //Parsing though array and adding to list if set.add returns false it means value is already available in set
        if (!set.add(arr[i])) {
            list.add(arr[i]);
        }
    }
    //Parsing though array and checking if each element is not available in set,then that is smallest number
    for (int i = 0; i < arr.length; i++) {
        if (!list.contains(arr[i])) {
            min = arr[i];
            break;
        }
    }
    Log.e(TAG, "firstNonRepeating: called===" + min);
    return min;
}

Answer 8

Try this :

int a[] = {1,2,3,4,5,1,2};
        
        for(int i=0; i<a.length;i++) {
            int count = 0;
            for(int j=0; j<a.length;j++) {
                if(a[i]==a[j] && i!=j) {
                    count++;
                    break;
                }
            }
                if(count == 0) {
                    System.out.println(a[i]);
                    break;              //To display first non repeating element
                }
        }

Answer 9

Using JS Object:

function nonRepeat_Using_Object(arr) {
    const data = arr.reduce((acc, val) => {
        if (!acc[val]) {
            acc[val] = 0;
        }
        acc[val]++;
        return acc;
    }, {});
    for (let i = 0; i < arr.length; i++) {
        if (data[arr[i]] === 1) {
            return arr[i];
        }
    }
}

Answer 10

实现此目的的另一种方法：您可以使用哈希图在第一次传递中存储整数计数，并在第二次传递中返回计数为 1 的第一个元素。