假设我们有一个二进制数,即数字n的表示。我们必须找到一个最小但大于n的数字的二进制表示形式,并且它也具有相同的数字0和1。因此,如果数字为1011(十进制为11),那么输出将为1101(13)。使用下一次排列计算可以找到此问题。让我们看一下获得想法的算法。
nextBin(bin)-
Begin len := length of the bin for i in range len-2, down to 1, do if bin[i] is 0 and bin[i+1] = 1, then exchange the bin[i] and bin[i+1] break end if done if i = 0, then there is no change, return otherwise j:= i + 2, k := len – 1 while j < k, do if bin[j] is 1 and bin[k] is 0, then exchange bin[j] and bin[k] increase j and k by 1 else if bin[i] is 0, then break else increase j by 1 end if done return bin End
#include <iostream>
using namespace std;
string nextBinary(string bin) {
int len = bin.size();
int i;
for (int i=len-2; i>=1; i--) {
if (bin[i] == '0' && bin[i+1] == '1') {
char ch = bin[i];
bin[i] = bin[i+1];
bin[i+1] = ch;
break;
}
}
if (i == 0)
"No greater number is present";
int j = i+2, k = len-1;
while (j < k) {
if (bin[j] == '1' && bin[k] == '0') {
char ch = bin[j];
bin[j] = bin[k];
bin[k] = ch;
j++;
k--;
}
else if (bin[i] == '0')
break;
else
j++;
}
return bin;
}
int main() {
string bin = "1011";
cout << "Binary value of next greater number = " << nextBinary(bin);
}输出结果
Binary value of next greater number = 1101