通过交换具有奇数差的相邻数字来最小化给定数字

给定一个长度为N的字符串S代表一个仅由字符'1'、'2' 和 '3' 组成的整数，任务是通过交换相邻字符（如果它们的绝对值）来找到可以从字符串中形成的最小数任何次数的差异都是奇数。

例子：

Input: S = “213123”
Output: 122313
Explanation: The integer representing the output string (i.e. 122313) is smaller than the input string. 
The swapping done are as follows (characters to be swapped are in bold) :
213123 -> 123123
123123 -> 123213
123213 -> 122313

Input: S = “122333”
Output: 122333

方法：这个问题可以通过使用贪心方法很容易地解决，使用以下观察：

1 and 3 cannot be swapped as their difference is 2 (which is even). 
1 and 2 can be swapped and so can be 2 and 3 as their difference is 1 (odd).

So keeping that in mind all the 2s can be placed at any position in the resultant string. The optimal position is to place all the 2s after the 1s which are situated at the starting of the string.

按照下面提到的步骤来实施观察：

• 计算字符串中出现的所有 2。
• 如果它们位于字符串的开头，则保持 1 组不变，即也将它们保持在结果字符串的开头。
• 在这组 1 之后添加字符串的所有 2。
• 最后，将剩余的 1 和 3 以与给定字符串中相同的顺序添加到结果字符串中（因为 1 和 3 不能交换）。

下面是上述方法的实现。

C++

// C++ program for above approach
 
#include 
using namespace std;
 
// Function to find minimum number
// corresponding to the given string
// by swapping adjacent numbers if
// their difference is odd
string findMinimumString(string S)
{
    // Finding length of given string
    int N = S.length();
 
    // Declaring variable to store
    // the first occurrence of 3
    int pos = -1;
 
    // Declaring variables to store
    // the count of 1 before the first 3
    // and count of all 2 in given string
    int count1 = 0, count2 = 0;
 
    // Traversing the string and counting the
    // 1s before first occurrence of 3
    // and count of 2s in whole string
    for (int i = 0; i < N; i++) {
        if (pos == -1 && S[i] == '1')
            count1++;
        else if (pos == -1 && S[i] == '3')
            pos = i;
        else if (S[i] == '2')
            count2++;
    }
 
    // Declaring empty string to store answer
    string answer = "";
 
    // Adding all occurrences of 1
    // before first occurrence of 3
    // to the answer
    while (count1--)
        answer += '1';
 
    // Adding all the occurrences of 2
    // into the answer
    while (count2--)
        answer += '2';
 
    // Adding the rest of the occurrences
    // of 1 and 3 to the answer in same order
    // as that in given string
    if (pos != -1) {
        while (pos < N) {
            if (S[pos] != '2')
                answer += S[pos];
            pos++;
        }
    }
 
    // Returning answer string
    return answer;
}
 
// Driver Code
int main()
{
    string S = "213123";
    string minimum_string
        = findMinimumString(S);
    cout << minimum_string;
    return 0;
}

Java

// JAVA program for above approach
import java.util.*;
class GFG
{
   
    // Function to find minimum number
    // corresponding to the given string
    // by swapping adjacent numbers if
    // their difference is odd
    public static String findMinimumString(String S)
    {
       
        // Finding length of given string
        int N = S.length();
 
        // Declaring variable to store
        // the first occurrence of 3
        int pos = -1;
 
        // Declaring variables to store
        // the count of 1 before the first 3
        // and count of all 2 in given string
        int count1 = 0, count2 = 0;
 
        // Traversing the string and counting the
        // 1s before first occurrence of 3
        // and count of 2s in whole string
        for (int i = 0; i < N; i++) {
            if (pos == -1 && S.charAt(i) == '1')
                count1++;
            else if (pos == -1 && S.charAt(i) == '3')
                pos = i;
            else if (S.charAt(i) == '2')
                count2++;
        }
 
        // Declaring empty string to store answer
        String answer = "";
 
        // Adding all occurrences of 1
        // before first occurrence of 3
        // to the answer
        while ((count1--) != 0)
            answer += '1';
 
        // Adding all the occurrences of 2
        // into the answer
        while ((count2--) != 0)
            answer += '2';
 
        // Adding the rest of the occurrences
        // of 1 and 3 to the answer in same order
        // as that in given string
        if (pos != -1) {
            while (pos < N) {
                if (S.charAt(pos) != '2')
                    answer += S.charAt(pos);
                pos++;
            }
        }
 
        // Returning answer string
        return answer;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "213123";
        String minimum_string = findMinimumString(S);
        System.out.print(minimum_string);
    }
}
 
// This code is contributed by Taranpreet

Python3

# Function to find minimum number
# corresponding to the given string
# by swapping adjacent numbers if
# their difference is odd
def findMinimumString(S) :
     
    # Finding length of given string
    N = len(S)
 
    # Declaring variable to store
    # the first occurrence of 3
    pos = -1
 
    # Declaring variables to store
    # the count of 1 before the first 3
    # and count of all 2 in given string
    count1 = 0
    count2 = 0
 
    # Traversing the string and counting the
    # 1s before first occurrence of 3
    # and count of 2s in whole string
    for i in range(N) :
        if (pos == -1 and S[i] == '1'):
            count1 += 1
        elif (pos == -1 and S[i] == '3'):
            pos = i
        elif (S[i] == '2') :
            count2 += 1
 
    # Declaring empty string to store answer
    answer = ""
 
    # Adding all occurrences of 1
    # before first occurrence of 3
    # to the answer
    while (count1) :
        answer += '1'
        count1 -= 1
 
    # Adding all the occurrences of 2
    # into the answer
    while (count2):
        answer += '2'
        count2 -= 1
 
    # Adding the rest of the occurrences
    # of 1 and 3 to the answer in same order
    # as that in given string
    if (pos != -1) :
        while (pos < N) :
            if (S[pos] != '2'):
                answer += S[pos]
            pos += 1
         
    # Returning answer string
    return answer
 
# Driver Code
S = "213123"
minimum_string = findMinimumString(S)
print(minimum_string)
 
# This code is contributed by sanjoy_62.

C#

// C# program for above approach
using System;
class GFG
{
 
  // Function to find minimum number
  // corresponding to the given string
  // by swapping adjacent numbers if
  // their difference is odd
  static String findMinimumString(string S)
  {
 
    // Finding length of given string
    int N = S.Length;
 
    // Declaring variable to store
    // the first occurrence of 3
    int pos = -1;
 
    // Declaring variables to store
    // the count of 1 before the first 3
    // and count of all 2 in given string
    int count1 = 0, count2 = 0;
 
    // Traversing the string and counting the
    // 1s before first occurrence of 3
    // and count of 2s in whole string
    for (int i = 0; i < N; i++) {
      if (pos == -1 && S[i] == '1')
        count1++;
      else if (pos == -1 && S[i] == '3')
        pos = i;
      else if (S[i] == '2')
        count2++;
    }
 
    // Declaring empty string to store answer
    string answer = "";
 
    // Adding all occurrences of 1
    // before first occurrence of 3
    // to the answer
    while ((count1--) != 0)
      answer += '1';
 
    // Adding all the occurrences of 2
    // into the answer
    while ((count2--) != 0)
      answer += '2';
 
    // Adding the rest of the occurrences
    // of 1 and 3 to the answer in same order
    // as that in given string
    if (pos != -1) {
      while (pos < N) {
        if (S[pos] != '2')
          answer += S[pos];
        pos++;
      }
    }
 
    // Returning answer string
    return answer;
  }
 
  // Driver Code
  public static void Main()
  {
    string S = "213123";
    string minimum_string = findMinimumString(S);
    Console.Write(minimum_string);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for above approach
 
    // Function to find minimum number
    // corresponding to the given string
    // by swapping adjacent numbers if
    // their difference is odd
    const findMinimumString = (S) => {
     
        // Finding length of given string
        let N = S.length;
 
        // Declaring variable to store
        // the first occurrence of 3
        let pos = -1;
 
        // Declaring variables to store
        // the count of 1 before the first 3
        // and count of all 2 in given string
        let count1 = 0, count2 = 0;
 
        // Traversing the string and counting the
        // 1s before first occurrence of 3
        // and count of 2s in whole string
        for (let i = 0; i < N; i++) {
            if (pos == -1 && S[i] == '1')
                count1++;
            else if (pos == -1 && S[i] == '3')
                pos = i;
            else if (S[i] == '2')
                count2++;
        }
 
        // Declaring empty string to store answer
        let answer = "";
 
        // Adding all occurrences of 1
        // before first occurrence of 3
        // to the answer
        while (count1--)
            answer += '1';
 
        // Adding all the occurrences of 2
        // into the answer
        while (count2--)
            answer += '2';
 
        // Adding the rest of the occurrences
        // of 1 and 3 to the answer in same order
        // as that in given string
        if (pos != -1) {
            while (pos < N) {
                if (S[pos] != '2')
                    answer += S[pos];
                pos++;
            }
        }
 
        // Returning answer string
        return answer;
    }
 
    // Driver Code
 
    let S = "213123";
    let minimum_string
        = findMinimumString(S);
    document.write(minimum_string);
 
// This code is contributed by rakeshsahni
 

输出

122313

时间复杂度： O(N)
辅助空间： O(N)