给定整数N ，任务是找到将N的二进制表示形式转换为回文式所需的最少位数。
例子：
Input: N = 12
Output: 2
Explanation:
Binary String representing 12 = “1100”.
To make “1100” a palindrome, convert the string to “0110”.
Therefore, minimum bits required to be flipped is 2.
Input: N = 7
Output: 0
Explanation:
Binary String representing 7 = 111, which is already a palindrome.
天真的方法：最简单的方法是检查每个可能的子集，即具有相同位数的回文集。
时间复杂度： O(N)
辅助空间： O(1)
高效方法：可以通过以下步骤优化上述方法：
- 首先，检查给定数字的二进制形式的长度。
- 在LSB指向两个指针，在MSB指向另一个指针。
- 现在，继续递减第一个指针，并递增第二个指针。
- 检查第一和第二指针位置的位是否相同。如果不是，请增加要更改的位数。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to calculate the
// length of the binary string
int check_length(int n)
{
// Length
int ans = 0;
while (n) {
// Right shift of n
n = n >> 1;
// Increment the length
ans++;
}
// Return the length
return ans;
}
// Function to check if the bit present
// at i-th position is a set bit or not
int check_ith_bit(int n, int i)
{
// Returns true if the bit is set
return (n & (1 << (i - 1)))
? true
: false;
}
// Function to count the minimum
// number of bit flips required
int no_of_flips(int n)
{
// Length of the binary form
int len = check_length(n);
// Number of flips
int ans = 0;
// Pointer to the LSB
int right = 1;
// Pointer to the MSB
int left = len;
while (right < left) {
// Check if the bits are equal
if (check_ith_bit(n, right)
!= check_ith_bit(n, left))
ans++;
// Decrementing the
// left pointer
left--;
// Incrementing the
// right pointer
right++;
}
// Returns the number of
// bits to flip.
return ans;
}
// Driver Code
int main()
{
int n = 12;
cout << no_of_flips(n);
return 0;
} Java
// Java program to implement
// the above approach
class GFG{
// Function to calculate the
// length of the binary string
static int check_length(int n)
{
// Length
int ans = 0;
while (n != 0)
{
// Right shift of n
n = n >> 1;
// Increment the length
ans++;
}
// Return the length
return ans;
}
// Function to check if the bit present
// at i-th position is a set bit or not
static boolean check_ith_bit(int n, int i)
{
// Returns true if the bit is set
return (n & (1<< (i - 1))) != 0 ? true : false;
}
// Function to count the minimum
// number of bit flips required
static int no_of_flips(int n)
{
// Length of the binary form
int len = check_length(n);
// Number of flips
int ans = 0;
// Pointer to the LSB
int right = 1;
// Pointer to the MSB
int left = len;
while (right < left)
{
// Check if the bits are equal
if (check_ith_bit(n, right) !=
check_ith_bit(n, left))
ans++;
// Decrementing the
// left pointer
left--;
// Incrementing the
// right pointer
right++;
}
// Returns the number of
// bits to flip.
return ans;
}
// Driver Code
public static void main(String[] args)
{
int n = 12;
System.out.println(no_of_flips(n));
}
}
// This code is contributed by rutvik_56Python3
# Python3 program to implement
# the above approach
# Function to calculate the
# length of the binary string
def check_length(n):
# Length
ans = 0
while (n):
# Right shift of n
n = n >> 1
# Increment the length
ans += 1
# Return the length
return ans
# Function to check if the bit present
# at i-th position is a set bit or not
def check_ith_bit(n, i):
# Returns true if the bit is set
if (n & (1 << (i - 1))):
return True
else:
return False
# Function to count the minimum
# number of bit flips required
def no_of_flips(n):
# Length of the binary form
ln = check_length(n)
# Number of flips
ans = 0
# Pointer to the LSB
right = 1
# Pointer to the MSB
left = ln
while (right < left):
# Check if the bits are equal
if (check_ith_bit(n, right) !=
check_ith_bit(n, left)):
ans += 1
# Decrementing the
# left pointer
left -= 1
# Incrementing the
# right pointer
right += 1
# Returns the number of
# bits to flip.
return ans
# Driver Code
n = 12
print(no_of_flips(n))
# This code is contributed by Shivam SinghC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to calculate the
// length of the binary string
static int check_length(int n)
{
// Length
int ans = 0;
while (n != 0)
{
// Right shift of n
n = n >> 1;
// Increment the length
ans++;
}
// Return the length
return ans;
}
// Function to check if the bit present
// at i-th position is a set bit or not
static bool check_ith_bit(int n, int i)
{
// Returns true if the bit is set
return (n & (1 << (i - 1))) != 0 ?
true : false;
}
// Function to count the minimum
// number of bit flips required
static int no_of_flips(int n)
{
// Length of the binary form
int len = check_length(n);
// Number of flips
int ans = 0;
// Pointer to the LSB
int right = 1;
// Pointer to the MSB
int left = len;
while (right < left)
{
// Check if the bits are equal
if (check_ith_bit(n, right) !=
check_ith_bit(n, left))
ans++;
// Decrementing the
// left pointer
left--;
// Incrementing the
// right pointer
right++;
}
// Returns the number of
// bits to flip.
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int n = 12;
Console.WriteLine(no_of_flips(n));
}
}
// This code is contributed by sapnasingh4991输出：
2
时间复杂度： O(log N)
辅助空间： O(1)