Leetcode/ numbers that appear only once

package com.xcrj;

import java.util.HashMap;
import java.util.Map;

/** *  The finger of the sword  Offer II 004.  A number that appears only once  *  Give you an array of integers  nums , Except that one element only appears   once   Outside , Every other element just happens to appear   Three times  . Please find and return the element that only appears once . *  Investigate ： *  Hash table statistics  *  Or operation splicing  result |= (1 << i); */
public class Solution4 {
    
    /** *  Ordinary brute force method   Traverse  */
    public int singleNumber(int[] nums) {
    
        int count = 0;
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
    
            result = nums[i];

            for (int j = 0; j < nums.length; j++) {
    
                if (result == nums[j]) {
    
                    count++;
                }
                if (count > 1) {
    
                    break;
                }
            }

            if (count == 1) {
    
                return result;
            }

            count = 0;
        }
        return result;
    }

    /** *  Hash table  */
    public int singleNumber2(int[] nums) {
    
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
    
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        // findFirst There may be many that only appear 1 The number of times 
        return map.entrySet().stream().filter(entry -> entry.getValue() == 1).findFirst().get().getKey();
    }

    /** *  In turn, the calculation appears 1 Every binary bit of the secondary element , Then use or operation to splice out 1 Sub elements  */
    public int singleNumber3(int[] nums) {
    
        int result = 0;
        // 32 because int yes 32 Bit 
        for (int i = 0; i < 32; ++i) {
    
            int total = 0;
            /** *  Find the number of all elements in the array i Sum of bits total *  Because only 3 Time or 1 Secondary elements  *  therefore total=3*m*1+3*n*0+0/1,m The first i Position as 1 Appearance 3 Number of secondary elements ,n The first i Position as 0 Appearance 3 Number of secondary elements , appear 1 Second element i Is it 0 perhaps 1 *  So there comes 1 Second element i position =total%3 */
            for (int num : nums) {
    
                total += ((num >> i) & 1);
            }
            //  It is known that   appear 1 Every bit of the secondary element   Splice through or operation to appear 1 Secondary elements .0 After moving left, it's still 0 No or operation splicing 
            if (total % 3 == 1) {
    
                result |= (1 << i);
            }
        }
        return result;
    }

    public static void main(String[] args) {
    
        Solution4 solution4 = new Solution4();
        System.out.println(solution4.singleNumber3(new int[]{
    1, 2, 1, 1}));
    }
}