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📜  通过交换相邻的偶数奇数对，可能的最小数量

📅  最后修改于: 2021-09-06 06:47:03             🧑  作者: Mango

给定一个数字字符串str ，任务是找到可以通过交换不同奇偶校验的相邻数字形成的最小整数。

例子：

方法：
我们可以观察到，在重复交换时，我们可以将str的偶数和奇数数字分成两个单独的块。由于不允许在块内交换(相同奇偶校验)，因此它们各自块中的数字顺序将与它们在字符串的出现顺序相同。因此，在将str分成两个单独的块之后，我们需要遍历这两个块并将当前指向的两个值中的最小值附加到答案中。在此操作之后生成的最后一个字符串，然后是删除前导 0(如果有)，就是必需的答案。

例子：
偶数块和奇数块在 836360 中出现的顺序分别是 {8, 6, 6, 0} 和 {3, 3}。
因此ANS形成的最小数目如下：

  1. ans = ans + min(8, 3) => ans = 3
  2. ans = ans + min(8, 3) => ans = 33
  3. 由于所有的奇数位都用完了，剩下的偶数位需要一一相加。
  4. 因此，所需的答案是 338660

下面是上述方法的实现：

C++
// C++ Program to find the
// smallest number possible
// by swapping adjacent digits
// of different parity
  
#include 
using namespace std;
  
// Function to return the
// smallest number possible
string findAns(string s)
{
    int digit;
  
    // Arrays to store odd and
    // even digits in the order
    // of their appearance in
    // the given string
    vector odd;
    vector even;
  
    // Insert the odd and
    // even digits
    for (auto c : s) {
        digit = c - '0';
        if (digit & 1)
            odd.push_back(digit);
        else
            even.push_back(digit);
    }
  
    // pointer to odd digit
    int i = 0;
    // pointer to even digit
    int j = 0;
  
    string ans = "";
  
    while (i < odd.size()
           and j < even.size()) {
  
        if (odd[i] < even[j])
            ans += (char)(odd[i++] + '0');
        else
            ans += (char)(even[j++] + '0');
    }
  
    // In case number of even and
    // odd digits are not equal
  
    // If odd digits are remaining
    while (i < odd.size())
        ans += (char)(odd[i++] + '0');
  
    // If even digits are remaining
    while (j < even.size())
        ans += (char)(even[j++] + '0');
  
    // Removal of leading 0's
    while (ans[0] == '0') {
        ans.erase(ans.begin());
    }
  
    return ans;
}
int main()
{
  
    string s = "894687536";
    cout << findAns(s);
    return 0;
}

Java
// Java program to find the smallest 
// number possible by swapping adjacent 
// digits of different parity
import java.util.*;
  
class GFG{
  
// Function to return the
// smallest number possible
static String findAns(String s)
{
    int digit;
  
    // Arrays to store odd and even 
    // digits in the order their 
    // appearance in the given String
    Vector odd =new Vector();
    Vector even = new Vector();
  
    // Insert the odd and
    // even digits
    for(char c : s.toCharArray())
    {
       digit = c - '0';
       if (digit % 2 == 1)
           odd.add(digit);
       else
           even.add(digit);
    }
  
    // Pointer to odd digit
    int i = 0;
      
    // Pointer to even digit
    int j = 0;
  
    String ans = "";
  
    while (i < odd.size() && j < even.size())
    {
        if (odd.get(i) < even.get(j))
            ans += (char)(odd.get(i++) + '0');
        else
            ans += (char)(even.get(j++) + '0');
    }
  
    // In case number of even and
    // odd digits are not equal
    // If odd digits are remaining
    while (i < odd.size())
        ans += (char)(odd.get(i++) + '0');
  
    // If even digits are remaining
    while (j < even.size())
        ans += (char)(even.get(j++) + '0');
  
    // Removal of leading 0's
    while (ans.charAt(0) == '0')
    {
        ans = ans.substring(1);
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    String s = "894687536";
      
    System.out.print(findAns(s));
}
}
  
// This code is contributed by 29AjayKumar

Python3
# Python3 Program to find the 
# smallest number possible 
# by swapping adjacent digits 
# of different parity 
  
# Function to return the 
# smallest number possible 
def findAns(s):
  
    # Arrays to store odd and 
    # even digits in the order 
    # of their appearance in 
    # the given string 
    odd = []
    even = [] 
  
    # Insert the odd and 
    # even digits 
          
    for c in s:
        digit = int(c)
        if (digit & 1):
            odd.append(digit)
        else:
            even.append(digit)
  
    # pointer to odd digit 
    i = 0
      
    # pointer to even digit 
    j = 0
  
    ans = ""
  
    while (i < len(odd) and j < len(even)):
          
        if (odd[i] < even[j]):
            ans += str(odd[i])
            i = i + 1
        else:
            ans += str(even[j])
            j = j + 1
  
    # In case number of even and 
    # odd digits are not equal 
  
    # If odd digits are remaining 
    while (i < len(odd)):
        ans += str(odd[i])
        i = i + 1
  
    # If even digits are remaining 
    while (j < len(even)):
        ans += str(even[j])
        j = j + 1
  
    # Removal of leading 0's 
    while (ans[0] == '0'):
        ans = ans[1:]
  
    return ans
  
# Driver Code
s = "894687536"
print(findAns(s))
  
# This code is contributed by yatin

C#
// C# program to find the smallest 
// number possible by swapping adjacent 
// digits of different parity
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to return the
// smallest number possible
static String findAns(String s)
{
    int digit;
  
    // Arrays to store odd and even 
    // digits in the order their 
    // appearance in the given String
    List odd = new List();
    List even = new List();
  
    // Insert the odd and
    // even digits
    foreach(char c in s.ToCharArray())
    {
        digit = c - '0';
        if (digit % 2 == 1)
            odd.Add(digit);
        else
            even.Add(digit);
    }
  
    // Pointer to odd digit
    int i = 0;
      
    // Pointer to even digit
    int j = 0;
  
    String ans = "";
  
    while (i < odd.Count && j < even.Count)
    {
        if (odd[i] < even[j])
            ans += (char)(odd[i++] + '0');
        else
            ans += (char)(even[j++] + '0');
    }
  
    // In case number of even and
    // odd digits are not equal
    // If odd digits are remaining
    while (i < odd.Count)
        ans += (char)(odd[i++] + '0');
  
    // If even digits are remaining
    while (j < even.Count)
        ans += (char)(even[j++] + '0');
  
    // Removal of leading 0's
    while (ans[0] == '0')
    {
        ans = ans.Substring(1);
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    String s = "894687536";
      
    Console.Write(findAns(s));
}
}
  
// This code is contributed by 29AjayKumar

输出： 
846869753

时间复杂度： O(N) ，其中 N 是给定字符串的大小。