下一个更大数字的二进制表示，具有相同的1和0

本文概述

建议：在继续解决方案之前, 请先在{IDE}上尝试使用你的方法。
C ++
Java
Python3
C#

给定一个表示正数n的二进制表示的二进制输入, 请找到大于n的最小数的二进制表示, 并且与n的二进制表示中的1和0相同。如果无法形成这样的数字, 请打印"没有更大的数字"。
即使是无符号long long int, 二进制输入可能也可能不适合。
例子：

Input : 10010Output : 10100Here n = (18)10 = (10010)2next greater = (20)10 = (10100)2Binary representation of 20 contains same number of1's and 0's as in 18.Input : 111000011100111110Output :111000011101001111

推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。这个问题简单地归结为找到给定字符串的下一个排列。我们可以找到next_permutation()输入的二进制数。
【下一个更大数字的二进制表示，具有相同的1和0】以下是查找二进制字符串中下一个置换的算法。

遍历二进制字符串bstr从右边。
遍历时找到第一个索引一世这样bstr [i] ='0'和bstr [i + 1] ='1'。
在索引" i"和" i + 1"处交换字符。
由于我们需要最小的下一个值, 因此请考虑索引中的子字符串我+2结束并移动所有1个在最后的子字符串中。

下面是上述步骤的实现。
C ++

// C++ program to find next permutation in a // binary string. #include < bits/stdc++.h> using namespace std; // Function to find the next greater number // with same number of 1's and 0's string nextGreaterWithSameDigits(string bnum) { int l = bnum.size(); int i; for ( int i=l-2; i> =1; i--) { // locate first 'i' from end such that // bnum[i]=='0' and bnum[i+1]=='1' // swap these value and break; if (bnum.at(i) == '0' & & bnum.at(i+1) == '1' ) { char ch = bnum.at(i); bnum.at(i) = bnum.at(i+1); bnum.at(i+1) = ch; break ; } }// if no swapping performed if (i == 0) "no greater number" ; // Since we want the smallest next value, // shift all 1's at the end in the binary // substring starting from index 'i+2' int j = i+2, k = l-1; while (j < k) { if (bnum.at(j) == '1' & & bnum.at(k) == '0' ) { char ch = bnum.at(j); bnum.at(j) = bnum.at(k); bnum.at(k) = ch; j++; k--; }// special case while swapping if '0' // occurs then break else if (bnum.at(i) == '0' ) break ; else j++; }// required next greater number return bnum; }// Driver program to test above int main() { string bnum = "10010" ; cout < < "Binary representation of next greater number = " < < nextGreaterWithSameDigits(bnum); return 0; }

Java

// Java program to find next permutation in a // binary string. class GFG {// Function to find the next greater number // with same number of 1's and 0's static String nextGreaterWithSameDigits( char [] bnum) { int l = bnum.length; int i; for (i = l - 2 ; i > = 1 ; i--) { // locate first 'i' from end such that // bnum[i]=='0' and bnum[i+1]=='1' // swap these value and break; if (bnum[i] == '0' & & bnum[i+ 1 ] == '1' ) { char ch = bnum[i]; bnum[i] = bnum[i+ 1 ]; bnum[i+ 1 ] = ch; break ; } }// if no swapping performed if (i == 0 ) System.out.println( "no greater number" ); // Since we want the smallest next value, // shift all 1's at the end in the binary // substring starting from index 'i+2' int j = i + 2 , k = l - 1 ; while (j < k) { if (bnum[j] == '1' & & bnum[k] == '0' ) { char ch = bnum[j]; bnum[j] = bnum[k]; bnum[k] = ch; j++; k--; }// special case while swapping if '0' // occurs then break else if (bnum[i] == '0' ) break ; else j++; }// required next greater number return String.valueOf(bnum); }// Driver program to test above public static void main(String[] args) { char [] bnum = "10010" .toCharArray(); System.out.println( "Binary representation of next greater number = " + nextGreaterWithSameDigits(bnum)); } }// This code contributed by Rajput-Ji

Python3

# Python3 program to find next permutation in a # binary string.# Function to find the next greater number # with same number of 1's and 0's def nextGreaterWithSameDigits(bnum): l = len (bnum) bnum = list (bnum) for i in range (l - 2 , 0 , - 1 ):# locate first 'i' from end such that # bnum[i]=='0' and bnum[i+1]=='1' # swap these value and break if (bnum[i] = = '0' and bnum[i + 1 ] = = '1' ): ch = bnum[i] bnum[i] = bnum[i + 1 ] bnum[i + 1 ] = ch break# if no swapping performed if (i = = 0 ): return "no greater number"# Since we want the smallest next value, # shift all 1's at the end in the binary # substring starting from index 'i+2' j = i + 2 k = l - 1 while (j < k): if (bnum[j] = = '1' and bnum[k] = = '0' ): ch = bnum[j] bnum[j] = bnum[k] bnum[k] = ch j + = 1 k - = 1# special case while swapping if '0' # occurs then break elif (bnum[i] = = '0' ): break else : j + = 1# required next greater number return bnum# Driver code bnum = "10010" print ( "Binary representation of next greater number = " , * nextGreaterWithSameDigits(bnum), sep = "")# This code is contributed by shubhamsingh10

// C# program to find next permutation in a // binary string. using System; class GFG {// Function to find the next greater number // with same number of 1's and 0's static String nextGreaterWithSameDigits( char [] bnum) { int l = bnum.Length; int i; for (i = l - 2; i > = 1; i--) { // locate first 'i' from end such that // bnum[i]=='0' and bnum[i+1]=='1' // swap these value and break; if (bnum[i] == '0' & & bnum[i+1] == '1' ) { char ch = bnum[i]; bnum[i] = bnum[i+1]; bnum[i+1] = ch; break ; } }// if no swapping performed if (i == 0) Console.WriteLine( "no greater number" ); // Since we want the smallest next value, // shift all 1's at the end in the binary // substring starting from index 'i+2' int j = i + 2, k = l - 1; while (j < k) { if (bnum[j] == '1' & & bnum[k] == '0' ) { char ch = bnum[j]; bnum[j] = bnum[k]; bnum[k] = ch; j++; k--; }// special case while swapping if '0' // occurs then break else if (bnum[i] == '0' ) break ; else j++; }// required next greater number return String.Join( "" , bnum); }// Driver code public static void Main(String[] args) { char [] bnum = "10010" .ToCharArray(); Console.WriteLine( "Binary representation of next greater number = " + nextGreaterWithSameDigits(bnum)); } }// This code is contributed by 29AjayKumar

输出如下：