使用for循环和。搜索值

Question

The question is this:

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Input: [4,1,2,1,2]
Output: 4

my code is:

public static int singleNumber(int[] nums) {
     int answer = 0;
        for (int i =0; i<nums.length-1; i++) {
            for(int j = i+1; j<nums.length; j++) {
                if(nums[i] != nums[j]) {
                 answer = nums[i];      //this should be where I am wrong.
                }
            }
        }
        return answer;
    }

I know that the output was 4 and then now it will be changed to 1. I'm trying to figure out how not to change the found value once found.

Answer 1

The logic is wrong - your inner loop finds every number that isn't the only number in the array.

I'd maintain a Set to track the numbers I've encountered. The first time you encounter a number, you add it to the Set . The second time you encounter it, you remove it from the Set . Once you're done going over the array, you'd have a Set with a single element, which is your answer:

public static int singleNumber(int[] nums) {
    Set<Integer> unique = new HashSet<>();
    for (int num : nums) {
        // add returns true if num is indeed new to unique
        if (!unique.add(num)) {
            unique.remove(num);
        }
    }

    return unique.iterator().next();
}

Answer 2

For this problem, i would to bitwise XOR of the numbers. The numbers that are equal will cancel one another and only single integer will be the final value.

public static int singleNumber(int[] nums) {
     int answer = 0;
        for (int i =0; i<nums.length; i++) {
           answer = answer ^ nums[i];
        }
        return answer;
 }

Answer 3

below changes to your method will give you the expected answer

public static int singleNumber(int[] nums) {

    int temp = 0;
    int answer = 0;

    for (int i = 0; i < nums.length; i++) {
        boolean flag = true;
        temp = nums[i];
        for (int j = 0; j < nums.length; j++) {
            if (temp == nums[j]) {
                if (i != j) {// if a match found then the loop will terminate
                    flag = false;
                    break;
                }
            }

        }
        if (flag == true) {
            answer = temp;
        }
    }
    return answer;
}

Answer 4

Here is another solution using Collectors.groupingBy from Java 8 :

public static int singleNumber(int[] nums) {
    return Arrays.stream(nums).boxed()
            .collect(Collectors.groupingBy(a -> a, Collectors.counting()))
            .entrySet().stream().filter(e -> e.getValue() == 1).findFirst().get().getKey();
}

the idea is :

• group by number of occurence
• then find the one which is repeated just one time

---

Note I assume that your array contain at least one element, else you can check the length before you search then throw an exception like this :

public static int singleNumber(int[] nums) throws IllegalArgumentException{
    if(nums.length == 0){
        throw new IllegalArgumentException("empty array");
    }
    return Arrays.stream(nums).boxed()
            .collect(Collectors.groupingBy(a -> a, Collectors.counting()))
            .entrySet().stream().filter(e -> e.getValue() == 1).findFirst().get().getKey();
}

---

More deeper, If you want to avoid the situation where there are multiple number repeated just one time you can use :

public static int singleNumber(int[] nums) throws IllegalArgumentException {
    if (nums.length == 0) {
        throw new IllegalArgumentException("empty array");
    }
    Map<Integer, Long> grouping = Arrays.stream(nums).boxed()
            .collect(Collectors.groupingBy(a -> a, Collectors.counting()));
    if (grouping.values().stream().filter(c -> c == 1).count() > 1) {
        throw new IllegalArgumentException("more than one element is repeated one time");
    }

    return grouping.entrySet().stream()
            .filter(e -> e.getValue() == 1).findFirst().get().getKey();
}

Answer 5

Here is a solution which uses ArrayList.indexOf and ArrayList.lastIndexOf. If they are same, you have your answer.

public static int singleNumber(int[] nums) {
    int answer = 0;
    //ArrayList<Integer> list = new ArrayList<Integer>(Arrays.asList(nums));
    ArrayList al = new ArrayList();
    for (int i =0; i < nums.length; i++) {
        al.add(nums[i]);
    }

    for (int i =0; i < nums.length; i++) {
        int test = nums[i];
        if(al.indexOf(test) == al.lastIndexOf(test)){
            answer = nums[i];
        }
    }
    return answer;
 }

Answer 6

Try this:

    int[] nums = new int[] {4,2,1,2,1};
         int answer = 0;
            for (int i =0; i<nums.length-1; i++) {
                int times = 0;
                int target = nums[i];
                for(int j : nums) {
                    if(j == target) {
                        times++;
                        if(times == 2) {
                            break;
                        }
                    }
                }
                if(times == 1) {
                    answer = target;
                    break;
                }
            }
            System.out.println(answer);

you have to go into every number and count how meny are in the array if there is only 1 you instantly stop the loop end set the answer