检查位是否在给定范围内处于备用模式

给定一个非负数n和两个值l和r 。问题是检查n是否在l到r的二进制表示形式中具有替代模式。在此，交替模式表示置位和未置位以交替顺序发生。这些位从右到左编号，即，最低有效位被认为在第一位置。

约束： 1 <= l <= r <= n的二进制表示形式的位数。

例子：

Input : n = 18, l = 1, r = 3
Output : Yes
(18)10 = (10010)2
The bits in the range 1 to 3 in the
binary representation of 18 are in
alternate order.

Input : n = 22, l = 2, r = 4
Output : No
(22)10 = (10110)2
The bits in the range 2 to 4 in the
binary representation of 22 are not in
alternate order.

算法：

bitsAreInAltPatrnInGivenTRange(n, l, r)
    
    // Shift bits of given number n by 
    // (l-1).
    num = n >> (l - 1)
    
    // Find first bit of range
    prev = num & 1

    // Traverse through remaining
    // bits. For every bit, check if
    // current bit is same as previous
    // or not.
    num = num >> 1    
    for i = 1 to (r - l)
        curr = num & 1
        
        if curr == prev then
            return false
        
        prev = curr
        num = num >> 1
    
    return true;

C++

// C++ implementation to check whether bits are in
// alternate pattern in the given range
#include 
  
using namespace std;
  
// function to check whether bits are in
// alternate pattern in the given range
bool bitsAreInAltPatrnInGivenTRange(unsigned int n,
                                    unsigned int l, 
                                    unsigned int r)
{
    unsigned int num, prev, curr;
  
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
  
    // get the bit at the last position in 'num'
    prev = num & 1;
  
    // right shift 'num' by 1
    num = num >> 1;
  
    // loop until there are bits in the given range
    for (int i = 1; i <= (r - l); i++) {
  
        // get the bit at the last position in 'num'
        curr = num & 1;
  
        // if true, then bits are not in alternate
        // pattern
        if (curr == prev)
            return false;
  
        // update 'prev'
        prev = curr;
  
        // right shift 'num' by 1
        num = num >> 1;
    }
  
    // bits are in alternate pattern in the
    // given range
    return true;
}
  
// Driver program to test above
int main()
{
    unsigned int n = 18;
    unsigned int l = 1, r = 3;
  
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation to check
// whether bits are in alternate
// pattern in the given range
class GFG
{
      
// function to check whether 
// bits are in alternate 
// pattern in the given range
static boolean bitsAreInAltPatrnInGivenTRange(int n, 
                                       int l, int r)
{
    int num, prev, curr;
  
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
  
    // get the bit at the
    // last position in 'num'
    prev = num & 1;
  
    // right shift 'num' by 1
    num = num >> 1;
  
    // loop until there are 
    // bits in the given range
    for (int i = 1; i <= (r - l); i++) 
    {
  
        // get the bit at the
        // last position in 'num'
        curr = num & 1;
  
        // if true, then bits are 
        // not in alternate pattern
        if (curr == prev)
            return false;
  
        // update 'prev'
        prev = curr;
  
        // right shift 'num' by 1
        num = num >> 1;
    }
  
    // bits are in alternate 
    // pattern in the given range
    return true;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 18;
    int l = 1, r = 3;
  
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by mits

Python3

# Python3 implementation to check
# whether bits are in alternate
# pattern in the given range
  
# function to check whether
# bits are in alternate 
# pattern in the given range
def bitsAreInAltPatrnInGivenTRange(n, l, r):
  
    # right shift n by (l - 1) bits
    num = n >> (l - 1);
  
    # get the bit at the 
    # last position in 'num'
    prev = num & 1;
  
    # right shift 'num' by 1
    num = num >> 1;
  
    # loop until there are 
    # bits in the given range
    for i in range(1,(r - l)):
  
        # get the bit at the 
        # last position in 'num'
        curr = num & 1;
  
        # if true, then bits are 
        # not in alternate pattern
        if (curr == prev):
            return False;
  
        # update 'prev'
        prev = curr;
  
        # right shift 'num' by 1
        num = num >> 1;
  
    # bits are in alternate 
    # pattern in the given range
    return True;
  
# Driver Code
if __name__ == "__main__":
    n = 18;
    l = 1;
    r = 3;
  
    if(bitsAreInAltPatrnInGivenTRange(n, l, r)):
        print("Yes");
    else:
        print("No");
  
# This Code is contributed by mits

C#

// C# implementation to check
// whether bits are in alternate
// pattern in the given range
using System;
  
class GFG
{
      
// function to check whether 
// bits are in alternate 
// pattern in the given range
static bool bitsAreInAltPatrnInGivenTRange(int n, 
                                    int l, int r)
{
    int num, prev, curr;
  
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
  
    // get the bit at the
    // last position in 'num'
    prev = num & 1;
  
    // right shift 'num' by 1
    num = num >> 1;
  
    // loop until there are 
    // bits in the given range
    for (int i = 1; i <= (r - l); i++) 
    {
  
        // get the bit at the
        // last position in 'num'
        curr = num & 1;
  
        // if true, then bits are 
        // not in alternate pattern
        if (curr == prev)
            return false;
  
        // update 'prev'
        prev = curr;
  
        // right shift 'num' by 1
        num = num >> 1;
    }
  
    // bits are in alternate 
    // pattern in the given range
    return true;
}
  
// Driver Code
static void Main()
{
    int n = 18;
    int l = 1, r = 3;
  
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by mits

PHP

> ($l - 1);
  
    // get the bit at the 
    // last position in 'num'
    $prev = $num & 1;
  
    // right shift 'num' by 1
    $num = $num >> 1;
  
    // loop until there are 
    // bits in the given range
    for ($i = 1; $i <= ($r - $l); $i++)
    {
  
        // get the bit at the 
        // last position in 'num'
        $curr = $num & 1;
  
        // if true, then bits are 
        // not in alternate pattern
        if ($curr == $prev)
            return false;
  
        // update 'prev'
        $prev = $curr;
  
        // right shift 'num' by 1
        $num = $num >> 1;
    }
  
    // bits are in alternate 
    // pattern in the given range
    return true;
}
  
// Driver Code
$n = 18;
$l = 1;
$r = 3;
  
if (bitsAreInAltPatrnInGivenTRange($n, $l, $r))
    echo "Yes";
else
    echo "No";
  
// This Code is contributed by mits
?>

输出：

Yes