Find the three numbers that appear once from the 103 numbers

1、 from 101 Find the one that appears once in the number 1 Number ？

eg：[5,7,6,7,5] find 6

//101
void demo01() {
    
    int arr[5] = {
     5,7,6,7,5 };
	
    int result;
    result = 0;
    for (int i = 0; i < 5; i++) {
    
        result ^= arr[i];
    }
    
    printf(" The number that appears once =%d\n", result);
}

By XOR operation , You can find the only number that appears ;

2、 from 102 Find the one that appears once in the number 2 Number ？

eg：[5,7,6,7,5,8] find 6,8

//101
int demo01(int arr[]) {
    
    //int arr[5] = { 5,7,6,7,5 };

    int result;
    result = 0;
    for (int i = 0; i < 6; i++) {
    
        result ^= arr[i];
    }
    
    //printf(" The number that appears once =%d\n", result);
    return result;
}

//102
void demo02() {
    
    
    int arr[6] = {
     5,7,6,7,5,8 };
    int flag = demo01(arr);
    int split_val = flag & (-flag);
    printf("%d\n", split_val);
    int res1, res2;
    res1 = 0; res2 = 0;    
    for (int i = 0; i < 6; i++) {
    
        if (arr[i]&split_val) {
    
            res1 ^= arr[i];
        }else {
    
            res2 ^= arr[i];
        }
    }
    printf(" First number =%d, The second number =%d\n", res1,res2);
}

1、 The difference between two different numbers can be found through all valued XORs ;
2、 adopt flag & (-flag) Operation can find the lowest value of all XOR results 1;
3、 Pass and lowest 1 And operation , You can separate two different numbers ; One is true , One is false ;

3、 from 103 Find the one that appears once in the number 3 Number ？

eg: [5,7,6,7,5,8,9] find 6,8,9

//102
void demo02(int arr[],int even,int odd) {
    
    
    int split_val = even & (-even);
    //printf("%d\n", split_val);
    int res1, res2;
    res1 = 0; res2 = 0;    
    for (int i = 0; i < 7; i++) {
    
        if (arr[i]&split_val) {
    
            res1 ^= arr[i];
        }else {
    
            res2 ^= arr[i];
        }
    }
    if (split_val& odd) {
    
        res1 ^= odd;
    }else {
    
        res2 ^= odd;
    }
    printf(" The second number =%d, The third number =%d\n", res1,res2);
}

//103
void demo03() {
    
    int arr[7] = {
     5,7,6,7,5,8,9 };
    int res1, res2,count1,count2;
    int flag;
    flag = 1;
    for (int i = 0; i < 32;i++) {
    
        res1=res2=count1=count2 = 0;
        for (int j = 0; j < 7; j++) {
    
            if (flag&arr[j]) {
    
                res1 ^= arr[j];
                count1++;
            }else {
    
                res2 ^= arr[j];
                count2++;
            }    
        }
        // Judge whether the even number is greater than 0
        if (count1 % 2 == 0 && res1!=0) {
    
            printf(" First number =%d", res2);
            demo02(arr, res1, res2);
            break;
        }
        else if (count2 % 2 == 0 && res2 != 0) {
    
            printf(" First number =%d", res1);
            demo02(arr, res2, res1);
            break;
        }
        // If not  0 3 
        flag << 1;
    }
}

1、 First , Law on the use of violence , from 1 Start , Every time I move left 1 Bit binary bit , To 32 Bit means moving 31 position ;
2、 Just divide the three separate numbers 2 Pile up , There are only two cases ; Odd heap ： Even heap = 3 ：0 or 1：2;
3、 Just judge that the XOR operation of the even heap is not 0, That is to say 1：2 This situation ; You find a singular number in the odd heap ;
4、 adopt demo02 function ; Thoughts and 2 Same question , But because of 103 Number , So the third singular will be assigned to one side ;
5、 We need to judge this number , Is it in the odd heap or the even heap , adopt split_val & odd The judgment is on that side , Can be in XOR once , You can eliminate this number ;