解法一：（异或XOR）相同的数字出现两次则归零

class Solution {
    public int singleNumber(int[] nums) {
        int result = 0;
        for(int num:nums){
            result ^= num;
        }
        return result;
    }
}

注意：

• 其他方法：HashList记录次数再查找数组；HashSet数组记录数字，出现第一次加入，出现第二次删除。他们时间空间复杂度都是O(n)，不符合题目要求。
• 多个元素只出现一次

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

class Solution {
    public List<Integer> findDuplicates(int[] nums) {
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            int num = Math.abs(nums[i]);
            int index = num - 1;
            if (nums[index] < 0) {
                result.add(num);
            } else {
                nums[index] *= -1;
            }
        }
        return result;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String input = scanner.nextLine();
        int[] nums = parseInput(input);
        Solution solution = new Solution();
        List<Integer> duplicates = solution.findDuplicates(nums);
        System.out.println(formatOutput(duplicates));
    }

    private static int[] parseInput(String input) {
        String cleaned = input.replaceAll("[\\[\\] ]", "");
        String[] parts = cleaned.split(",");
        int[] nums = new int[parts.length];
        for (int i = 0; i < parts.length; i++) {
            nums[i] = Integer.parseInt(parts[i]);
        }
        return nums;
    }

    private static String formatOutput(List<Integer> list) {
        StringBuilder sb = new StringBuilder("[");
        for (int i = 0; i < list.size(); i++) {
            sb.append(list.get(i));
            if (i < list.size() - 1) {
                sb.append(",");
            }
        }
        sb.append("]");
        return sb.toString();
    }
}