给定数字，则打印具有相同数字的下一个最高数字

Question

given a number we have to find next highest number and previous number which has same number of 1 bits in their binary representation

my solution is a brute force approach ie incrementing the number and counting number of ones if it has same number of 1's then print it

for previous number same approach as above one

is there any optimised solution instead of checking all the numbers?

Answer 1

To get the next highest number, simply search for the first instance of 01 , starting on the right and change this to 10 . Then collapse all the 1 s to the right of the swapped bits into their least significant positions.

Example:

 ||
00110000 // 0x30 in hex

After swap:

 ||
01010000 // 0x50 in hex

Next, collapse all the 1s to the right of the swapped bits into the least significant positions:

 ||---->
01000001 // 0x41 in hex

To get the previous number, search for the first instance of 10 starting from the right, and replace it with 01 . Then collapse all the 0 s after the swapped bits in to the least significant positions (alternatively, collapse all the 1 s after the swapped bits into their most significant positions).

Example:

    ||
01001001 // 0x48 in hex

After swap:

    ||
01000101 // 0x45 in hex

Next, collapse all the 0s to the right of the swapped bits into the least significant positions:

    ||->
01000110 // 0x46 in hex

Here is a short program that will display both the next higher and lower numbers.

#include <iostream>
#include <bitset>
#include <stdlib.h>

using namespace std;
#define NUMSIZE 8       // The number of bits to display

// Finds the first and second values, swaps them, and collapses.
void swap_and_collapse(bitset<NUMSIZE>& num, bool v1, bool v2, bool c) {
  // Find the v1 immediately followed by v2 in the number.
  int insert_pos = 0;
  for (int i=1; i<NUMSIZE; i++) {
    if ((num.test(i-1) == v2) && (num.test(i) == v1)) {
      // Found them.  Swap the values.
      num.flip(i-1); num.flip(i);
      break;
    } else {
      // Move any c bits to the beginning.
      if (num.test(i-1) == c) {
        num.flip(i-1);
        num.flip(insert_pos++);
      } // if (num.test(i-1) == c) 
    } // if ((num.test(i-1) == v2) && (num.test(i) == v1)) 
  } // for (int i=0; i<16; i++)
} // swap_and_collapse

int main(int argc, char* argv[]) {
  // Get the number from the command line and display in binary.
  int value = atoi(argv[1]);
  bitset<NUMSIZE> orig  (value);
  cout << "Original Number " << orig.to_ulong() << " is " << orig << endl;

  // Find the next higher number.
  bitset<NUMSIZE> higher(value);
  swap_and_collapse(higher, 0, 1, 1);
  // Find the next lower number.
  bitset<NUMSIZE> lower (value);
  swap_and_collapse(lower,  1, 0, 0);

  cout << "Higher number   " << higher.to_ulong() << " is " << higher << endl;
  cout << "Lower number    " <<  lower.to_ulong() << " is " <<  lower << endl;
} // main

Answer 2

here is the code to find the next highest number with equal number of ones

#include<iostream>
using namespace std;
int main()
{
int c;
cin>>c;
int n=c;
int c1=0;
while(((c&1)==0) && (c!=0) )
{
    c1++;
    c=c>>1;
}
cout<<"the no of zereos"<<c1<<endl;
cout<<c<<endl;
int c2=0;
while((c&1)==1)
{
    c2++;
    c=c>>1;
}
cout<<"no of ones"<<c2<<endl;
cout<<c;
int p=c1+c2;
n=n|(1<<p);
int a=1<<p;
int b=a-1;
cout<<endl<<b;
int mask=~b;
n=n&mask;
a=1<<(c2-1);
b=a-1;
n=n|b;
cout<<"\n"<<a;
cout<<"\n"<<n;

  }