生成字典上最小的字符串0、1 和 2，并允许相邻交换
给定一个仅包含字符0 、 1和2的字符串str ，您可以交换任意两个相邻（连续）字符0和1或任意两个相邻（连续）字符1和2 。任务是通过使用这些交换任意次数来获得最小可能的（按字典顺序）字符串。
例子：
Input: str = “100210”
Output: 001120
We can swap 0 and 1 OR we can swap 1 and 2. Swapping 0 and 2 is not allowed. All the swaps can happen for adjacent only.
Input: str = “2021”
Output: 1202
Note that 0 and 2 cannot be swapped
方法：您可以将所有1一起打印，因为1可以与其他字符交换，而0和2不能交换，因此所有0和2将遵循与原始字符串相同的顺序。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the required string
void printString(string str, int n)
{
// count number of 1s
int ones = 0;
for (int i = 0; i < n; i++)
if (str[i] == '1')
ones++;
// To check if the all the 1s
// have been used or not
bool used = false;
for (int i = 0; i < n; i++) {
if (str[i] == '2' && !used) {
used = 1;
// Print all the 1s if any 2 is encountered
for (int j = 0; j < ones; j++)
cout << "1";
}
// If str[i] = 0 or str[i] = 2
if (str[i] != '1')
cout << str[i];
}
// If 1s are not printed yet
if (!used)
for (int j = 0; j < ones; j++)
cout << "1";
}
// Driver code
int main()
{
string str = "100210";
int n = str.length();
printString(str, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to print the required string
static void printString(char[] str, int n)
{
// count number of 1s
int ones = 0;
for (int i = 0; i < n; i++)
if (str[i] == '1')
ones++;
// To check if the all the 1s
// have been used or not
boolean used = false;
for (int i = 0; i < n; i++)
{
if (str[i] == '2' && !used)
{
used = true;
// Print all the 1s if any 2 is encountered
for (int j = 0; j < ones; j++)
System.out.print("1");
}
// If str[i] = 0 or str[i] = 2
if (str[i] != '1')
System.out.print(str[i]);
}
// If 1s are not printed yet
if (!used)
for (int j = 0; j < ones; j++)
System.out.print("1");
}
// Driver code
public static void main(String[] args)
{
String str = "100210";
int n = str.length();
printString(str.toCharArray(), n);
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 implementation of the approach
# Function to print the required string
def printString(Str1, n):
# count number of 1s
ones = 0
for i in range(n):
if (Str1[i] == '1'):
ones += 1
# To check if the all the 1s
# have been used or not
used = False
for i in range(n):
if (Str1[i] == '2' and used == False):
used = 1
# Print all the 1s if any 2 is encountered
for j in range(ones):
print("1", end = "")
# If Str1[i] = 0 or Str1[i] = 2
if (Str1[i] != '1'):
print(Str1[i], end = "")
# If 1s are not printed yet
if (used == False):
for j in range(ones):
print("1", end = "")
# Driver code
Str1 = "100210"
n = len(Str1)
printString(Str1, n)
# This code is contributed
# by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the required string
static void printString(char[] str, int n)
{
// count number of 1s
int ones = 0;
for (int i = 0; i < n; i++)
if (str[i] == '1')
ones++;
// To check if the all the 1s
// have been used or not
bool used = false;
for (int i = 0; i < n; i++)
{
if (str[i] == '2' && !used)
{
used = true;
// Print all the 1s if any 2 is encountered
for (int j = 0; j < ones; j++)
Console.Write("1");
}
// If str[i] = 0 or str[i] = 2
if (str[i] != '1')
Console.Write(str[i]);
}
// If 1s are not printed yet
if (!used)
for (int j = 0; j < ones; j++)
Console.Write("1");
}
// Driver code
public static void Main(String[] args)
{
String str = "100210";
int n = str.Length;
printString(str.ToCharArray(), n);
}
}
// This code has been contributed by 29AjayKumarPHP
Javascript
输出：
001120