给定一个非负数 N .问题是在二进制表示中设置最右边的未设置位 N .如果没有未设置的位，则保持数字不变。
null
例如：
Input : 21Output : 23(21)10 = (10101)2Rightmost unset bit is at position 2(from right) as highlighted in the binary representation of 21.(23)10 = (10111)2The bit at position 2 has been set.Input : 15Output : 15
方法： 以下是步骤：
- 如果 N =0，返回1。
- 如果所有的 N 已设置，返回n.参考 这 邮递
- 否则对给定的数字执行按位not（相当于1的补码的运算）。顺其自然 号码 =~n。
- 拿到 最右边设定位的位置 属于 号码 .让这个职位 销售时点情报系统 .
- 回来 （1< .
C++
// C++ implementation to set the rightmost unset bit #include <bits/stdc++.h> using namespace std; // function to find the position // of rightmost set bit int getPosOfRightmostSetBit( int n) { return log2(n&-n)+1; } int setRightmostUnsetBit( int n) { // if n = 0, return 1 if (n == 0) return 1; // if all bits of 'n' are set if ((n & (n + 1)) == 0) return n; // position of rightmost unset bit in 'n' // passing ~n as argument int pos = getPosOfRightmostSetBit(~n); // set the bit at position 'pos' return ((1 << (pos - 1)) | n); } // Driver program to test above int main() { int n = 21; cout << setRightmostUnsetBit(n); return 0; } |
JAVA
// Java implementation to set // the rightmost unset bit class GFG { // function to find the position // of rightmost set bit static int getPosOfRightmostSetBit( int n) { return ( int )((Math.log10(n & -n)) / (Math.log10( 2 ))) + 1 ; } static int setRightmostUnsetBit( int n) { // if n = 0, return 1 if (n == 0 ) return 1 ; // if all bits of 'n' are set if ((n & (n + 1 )) == 0 ) return n; // position of rightmost unset bit in 'n' // passing ~n as argument int pos = getPosOfRightmostSetBit(~n); // set the bit at position 'pos' return (( 1 << (pos - 1 )) | n); } // Driver code public static void main(String arg[]) { int n = 21 ; System.out.print(setRightmostUnsetBit(n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 implementation to # set the rightmost unset bit import math # function to find the position # of rightmost set bit def getPosOfRightmostSetBit(n): return int (math.log2(n& - n) + 1 ) def setRightmostUnsetBit(n): # if n = 0, return 1 if (n = = 0 ): return 1 # if all bits of 'n' are set if ((n & (n + 1 )) = = 0 ): return n # position of rightmost unset bit in 'n' # passing ~n as argument pos = getPosOfRightmostSetBit(~n) # set the bit at position 'pos' return (( 1 << (pos - 1 )) | n) # Driver code n = 21 print (setRightmostUnsetBit(n)) # This code is contributed # by Anant Agarwal. |
C#
// C# implementation to set // the rightmost unset bit using System; class GFG{ // Function to find the position // of rightmost set bit static int getPosOfRightmostSetBit( int n) { return ( int )((Math.Log10(n & -n)) / (Math.Log10(2))) + 1; } static int setRightmostUnsetBit( int n) { // If n = 0, return 1 if (n == 0) return 1; // If all bits of 'n' are set if ((n & (n + 1)) == 0) return n; // Position of rightmost unset bit in 'n' // passing ~n as argument int pos = getPosOfRightmostSetBit(~n); // Set the bit at position 'pos' return ((1 << (pos - 1)) | n); } // Driver code public static void Main(String []arg) { int n = 21; Console.Write(setRightmostUnsetBit(n)); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript implementation to set // the rightmost unset bit // function to find the position // of rightmost set bit function getPosOfRightmostSetBit(n) { return ((Math.log10(n & -n)) / (Math.log10(2))) + 1; } function setRightmostUnsetBit(n) { // if n = 0, return 1 if (n == 0) return 1; // if all bits of 'n' are set if ((n & (n + 1)) == 0) return n; // position of rightmost unset bit in 'n' // passing ~n as argument let pos = getPosOfRightmostSetBit(~n); // set the bit at position 'pos' return ((1 << (pos - 1)) | n); } // Driver code let n = 21; document.write(setRightmostUnsetBit(n)); // This code is contributed by unknown2108 </script> |
输出：
23
替代实施 这个想法是使用 整数Tobinarysting（）
JAVA
import java.io.*; import java.util.Scanner; public class SetMostRightUnsetBit { public static void main(String args[]) { int a = 21 ; setMostRightUnset(a); } private static void setMostRightUnset( int a) { // will get a number with all set bits except the // first set bit int x = a ^ (a - 1 ); System.out.println(Integer.toBinaryString(x)); // We reduce it to the number with single 1's on // the position of first set bit in given number x = x & a; System.out.println(Integer.toBinaryString(x)); // Move x on right by one shift to make OR // operation and make first rightest unset bit 1 x = x >> 1 ; int b = a | x; System.out.println( "before setting bit " + Integer.toBinaryString(a)); System.out.println( "after setting bit " + Integer.toBinaryString(b)); } } |
Python3
def setMostRightUnset(a): # Will get a number with all set # bits except the first set bit x = a ^ (a - 1 ) print ( bin (x)[ 2 :]) # We reduce it to the number with # single 1's on the position of # first set bit in given number x = x & a print ( bin (x)[ 2 :]) # Move x on right by one shift to # make OR operation and make first # rightest unset bit 1 x = x >> 1 b = a | x print ( "before setting bit " , bin (a)[ 2 :]) print ( "after setting bit " , bin (b)[ 2 :]) # Driver Code if __name__ = = '__main__' : a = 21 setMostRightUnset(a) # This code is contributed by mohit kumar 29 |
C#
using System; class SetMostRightUnsetBit{ // Driver Code public static void Main(String []args) { int a = 21; setMostRightUnset(a); } private static void setMostRightUnset( int a) { // will get a number with all set bits // except the first set bit int x = a ^ (a - 1); Console.WriteLine(Convert.ToString(x)); // We reduce it to the number with single 1's on // the position of first set bit in given number x = x & a; Console.WriteLine(Convert.ToString(x)); // Move x on right by one shift to make OR // operation and make first rightest unset bit 1 x = x >> 1; int b = a | x; Console.WriteLine( "before setting bit " + Convert.ToString(a, 2)); Console.WriteLine( "after setting bit " + Convert.ToString(b, 2)); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> function setMostRightUnset(a) { // will get a number with all set bits except the // first set bit let x = a ^ (a - 1); document.write((x >>> 0).toString(2)+ "<br>" ); // We reduce it to the number with single 1's on // the position of first set bit in given number x = x & a; document.write((x >>> 0).toString(2)+ "<br>" ); // Move x on right by one shift to make OR // operation and make first rightest unset bit 1 x = x >> 1; let b = a | x; document.write( "before setting bit " + (a >>> 0).toString(2)+ "<br>" ); document.write( "after setting bit " + (b >>> 0).toString(2)+ "<br>" ); } let a = 21; setMostRightUnset(a); // This code is contributed by patel2127 </script> |
输出：
11before setting bit 10101after setting bit 10101
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