Daily question 1: the number of numbers in the array 2

link : The number of occurrences of numbers in an array 2

This question is another version of the previous blog , The link to the previous one is below ：

link : The number of occurrences of numbers in an array 1

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The difference between this question and the previous one is that the number of times it appears here is 3 And then 1 Time of , So the XOR method is not very good , We can find another way .
We want to , Since only one number appears once in this array , The others are three times , Then use an array to make these numbers appear three times , Count and add each of their bits , You will find that the number of each bit in the statistical array will be 3 Multiple , If there is another number that appears once , Then he added it to a certain binary digit , Will make the number of a bit in this array into a module 3 more than 1, Then we can find out that the number is 1 Hexadecimal digits of , Finally, the number is calculated by binary operation . in general ：

Count all the numbers in the array , From 1 Position to the first 32 How many decimal digits are there 1, Then find the module in the array 3 more than 1 Number of digits , The binary bit of this once occurring number is 1 Number of digits .

int singleNumber(int* nums, int numsSize){
    
    // Statistics first , So let's start with an array 
    int arr[32] = {
    0};
    for(int i = 0; i < numsSize; i++)
    {
    
        for(int j = 0; j < 32; j++)
        {
    
            if(((nums[i]>>j) & 1) == 1)
            {
    
                arr[j] += 1;
            }
        }
    }

    // Let's see who's here once 
    int n = 0;
    for(int i = 0; i < 32; i++)
    {
    
        if((arr[i] % 3) == 1)
        {
    
            // Judge whether it is the No 0 Position as 1 The situation of 
            if(i == 0)
            {
    
                n += 1;
            }
            else
            {
    
                n += pow(2, i);
            }
        }
    }

    return n;
}