[leetcode -- the second day of introduction to programming ability] operator (the number of bit 1 / the difference between the sum of the products of integers)

subject ： position 1 The number of

Catalog

subject ： position 1 The number of

   analysis ：

  Solution one code ：

Solution II code ：

subject ： The difference between the sum of the products of integers

  analysis ：

Code ：

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Write a function , Input is an unsigned integer （ In the form of a binary string ）, Returns the number of digits in its binary expression as '1' The number of （ Also known as Han Ming weight ）.

Tips ：

Please note that , In some languages （ Such as Java） in , There is no unsigned integer type . under these circumstances , Both input and output will be specified as signed integer types , And should not affect your implementation , Because whether integers are signed or unsigned , Its internal binary representation is the same .
stay Java in , The compiler uses binary complement notation to represent signed integers . therefore , Above   Example 3  in , The input represents a signed integer -3.
 

Example 1：

Input ：00000000000000000000000000001011
Output ：3
explain ： Binary string of input 00000000000000000000000000001011  in , There are three of them '1'.
Example 2：

Input ：00000000000000000000000010000000
Output ：1
explain ： Binary string of input 00000000000000000000000010000000  in , There is one in all for '1'.
Example 3：

Input ：11111111111111111111111111111101
Output ：31
explain ： Binary string of input 11111111111111111111111111111101 in , share 31 Position as '1'.
 

Tips ：

The input must be of length 32 Of Binary string .
 

Advanced ：

If you call this function more than once , How will you optimize your algorithm ？

   analysis ：

I use it. Integer.BinaryString Convert the decimal system into binary system, and then convert it into a string. Loop through and intercept each character to determine whether it is 1 The way of accumulation , I think this method is cumbersome . So I saw the solution of other big guys and used Integer.bitCount();bigCount What is it? ？      Let's take a look at the original code

public static int bitCount(int i) {
        // HD, Figure 5-2
        i = i - ((i >>> 1) & 0x55555555);
        i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
        i = (i + (i >>> 4)) & 0x0f0f0f0f;
        i = i + (i >>> 8);
        i = i + (i >>> 16);
        return i & 0x3f;
    }

bigCount()： Return to binary complement 1 Number of bits . Using this function, you can directly find the number of one in binary ！ At present, I don't understand the bit operation , Welcome the boss to provide valuable guidance .

  Solution 1 Code ：

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
return Integer.bitCount(n);
    }
}

Solution 2 Code ：

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
  String s=Integer.toBinaryString(n);
  int sum=0;
  for(int i=0;i<s.length();i++){
      if(s.charAt(i)=='1')
      sum++;
  }
  return sum;
    }
}

subject ： The difference between the sum of the products of integers

Give you an integer  n, Please help calculate and return the integer 「 The product of your numbers 」 And 「 The sum of your numbers 」 Difference .

Example 1：

Input ：n = 234
Output ：15 
explain ：
Product of numbers = 2 * 3 * 4 = 24 
The sum of your numbers = 2 + 3 + 4 = 9 
result = 24 - 9 = 15
Example 2：

Input ：n = 4421
Output ：21
explain ： 
Product of numbers = 4 * 4 * 2 * 1 = 32 
The sum of your numbers = 4 + 4 + 2 + 1 = 11 
result = 32 - 11 = 21

Tips ：

1 <= n <= 10^5

  analysis ：

We can use %10 Take the end ,/10 Cut the end of the method to get the number of each position and then multiply 、 Add it up .

Code ：

class Solution {
    public int subtractProductAndSum(int n) {
     int i=1;
     int j=0;
     int k=n;
     while(k!=0){
      i*=k%10;
      j+=k%10;
      k/=10;
     }
     return i-j;
    }
}

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