给定一个非负整数 N .问题是增加 N 通过操纵 N . 例如：
null
Input : 6Output : 7Input : 15Output : 16
方法： 以下是步骤：
- 获取最右边未设置位的位置 属于 N .让这个职位 K .
- 设置n的第k位 .
- 切换最后的k-1位 属于 N .
- 最后，返回 N .
C++
// C++ implementation to increment a number // by one by manipulating the bits #include <bits/stdc++.h> using namespace std; // function to find the position // of rightmost set bit int getPosOfRightmostSetBit( int n) { return log2(n & -n); } // function to toggle the last m bits unsigned int toggleLastKBits(unsigned int n, unsigned int k) { // calculating a number 'num' having 'm' bits // and all are set unsigned int num = (1 << k) - 1; // toggle the last m bits and return the number return (n ^ num); } // function to increment a number by one // by manipulating the bits unsigned int incrementByOne(unsigned int n) { // get position of rightmost unset bit // if all bits of 'n' are set, then the // bit left to the MSB is the rightmost // unset bit int k = getPosOfRightmostSetBit(~n); // kth bit of n is being set by this operation n = ((1 << k) | n); // from the right toggle all the bits before the // k-th bit if (k != 0) n = toggleLastKBits(n, k); // required number return n; } // Driver program to test above int main() { unsigned int n = 15; cout << incrementByOne(n); return 0; } |
JAVA
// Java implementation to increment a number // by one by manipulating the bits import java.io.*; import java.util.*; class GFG { // function to find the position // of rightmost set bit static int getPosOfRightmostSetBit( int n) { return ( int )(Math.log(n & -n) / Math.log( 2 )); } // function to toggle the last m bits static int toggleLastKBits( int n, int k) { // calculating a number 'num' having // 'm' bits and all are set int num = ( 1 << k) - 1 ; // toggle the last m bits and return // the number return (n ^ num); } // function to increment a number by one // by manipulating the bits static int incrementByOne( int n) { // get position of rightmost unset bit // if all bits of 'n' are set, then the // bit left to the MSB is the rightmost // unset bit int k = getPosOfRightmostSetBit(~n); // kth bit of n is being set // by this operation n = (( 1 << k) | n); // from the right toggle all // the bits before the k-th bit if (k != 0 ) n = toggleLastKBits(n, k); // required number return n; } // Driver Program public static void main (String[] args) { int n = 15 ; System.out.println(incrementByOne(n)); } } // This code is contributed by Gitanjali. |
Python 3
# python 3 implementation # to increment a number # by one by manipulating # the bits import math # function to find the # position of rightmost # set bit def getPosOfRightmostSetBit(n) : return math.log2(n & - n) # function to toggle the last m bits def toggleLastKBits(n, k) : # calculating a number # 'num' having 'm' bits # and all are set num = ( 1 << ( int )(k)) - 1 # toggle the last m bits and # return the number return (n ^ num) # function to increment # a number by one by # manipulating the bits def incrementByOne(n) : # get position of rightmost # unset bit if all bits of # 'n' are set, then the bit # left to the MSB is the # rightmost unset bit k = getPosOfRightmostSetBit(~n) # kth bit of n is being # set by this operation n = (( 1 << ( int )(k)) | n) # from the right toggle # all the bits before the # k-th bit if (k ! = 0 ) : n = toggleLastKBits(n, k) # required number return n # Driver program n = 15 print (incrementByOne(n)) # This code is contributed # by Nikita Tiwari. |
C#
// C# implementation to increment a number // by one by manipulating the bits using System; class GFG { // function to find the position // of rightmost set bit static int getPosOfRightmostSetBit( int n) { return ( int )(Math.Log(n & -n) / Math.Log(2)); } // function to toggle the last m bits static int toggleLastKBits( int n, int k) { // calculating a number 'num' having // 'm' bits and all are set int num = (1 << k) - 1; // toggle the last m bits and return // the number return (n ^ num); } // function to increment a number by one // by manipulating the bits static int incrementByOne( int n) { // get position of rightmost unset bit // if all bits of 'n' are set, then the // bit left to the MSB is the rightmost // unset bit int k = getPosOfRightmostSetBit(~n); // kth bit of n is being set // by this operation n = ((1 << k) | n); // from the right toggle all // the bits before the k-th bit if (k != 0) n = toggleLastKBits(n, k); // required number return n; } // Driver Program public static void Main () { int n = 15; Console.WriteLine(incrementByOne(n)); } } // This code is contributed by Sam007. |
PHP
<?php // PHP implementation to increment a number // by one by manipulating the bits // function to find the position // of rightmost set bit function getPosOfRightmostSetBit( $n ) { $t = $n & - $n ; return log( $t , 2); } // function to toggle the last m bits function toggleLastKBits( $n , $k ) { // calculating a number 'num' // having 'm' bits and all are set $num = (1 << $k ) - 1; // toggle the last m bits and // return the number return ( $n ^ $num ); } // function to increment a number by one // by manipulating the bits function incrementByOne( $n ) { // get position of rightmost unset bit // if all bits of 'n' are set, then the // bit left to the MSB is the rightmost // unset bit $k = getPosOfRightmostSetBit(~ $n ); // kth bit of n is being set // by this operation $n = ((1 << $k ) | $n ); // from the right toggle all the // bits before the k-th if ( $k != 0) $n = toggleLastKBits( $n , $k ); // required number return $n ; } // Driver code $n = 15; echo incrementByOne( $n ); // This code is contributed by Mithun Kumar ?> |
Javascript
<script> // JavaScript program to increment a number // by one by manipulating the bits // function to find the position // of rightmost set bit function getPosOfRightmostSetBit(n) { return (Math.log(n & -n) / Math.log(2)); } // function to toggle the last m bits function toggleLastKBits(n, k) { // calculating a number 'num' having // 'm' bits and all are set let num = (1 << k) - 1; // toggle the last m bits and return // the number return (n ^ num); } // function to increment a number by one // by manipulating the bits function incrementByOne(n) { // get position of rightmost unset bit // if all bits of 'n' are set, then the // bit left to the MSB is the rightmost // unset bit let k = getPosOfRightmostSetBit(~n); // kth bit of n is being set // by this operation n = ((1 << k) | n); // from the right toggle all // the bits before the k-th bit if (k != 0) n = toggleLastKBits(n, k); // required number return n; } // Driver code let n = 15; document.write(incrementByOne(n)); </script> |
输出：
16
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