重新排列字符串

小鱼儿 2023-03-05 09:40 5阅读 0赞

**Problem statement:**

**问题陈述：**

Given a string containing uppercase alphabets and integer digits (from 0 to 9), the task is to print the alphabets in the order followed by the sum of digits.

给定一个包含大写字母和整数(从0到9)的字符串，任务是按字母的顺序打印，然后跟着数字的总和。

**Example:**

**例：**

Input:
        XY3BFA9KA2
    
        Output:
        AABFKXY14

**Solution**

**解**

Problem can be easily solved with help of a **map**. (Read more: [maps in C++ STL)][maps in C_ STL]

借助**地图**可以轻松解决问题。 (了解更多： [C ++ STL中的地图)][maps in C_ STL]

**Pre-requisite:** Input string s

**前提条件：**输入字符串s

**Algorithm:**

**算法：**

1.  Declare sum=0 to add integers of the string.
    
    声明sum = 0以添加字符串的整数。
2.  Declare map<char,int>m to tore characters of the string with respective frequencies;
    
    声明map <char，int> m以相应的频率撕裂字符串中的字符；
3.      For i=0:s.length()-1
            IF(s[i]>='0' && s[i]<='9') //current character is digit 
                sum+=s[i]-'0'; //add to sum//s[i]-'0'=actual numeric value
            Else //it's a alphabetical character
                m[s[i]]++; //store character with its frequency
            END IF
        END FOR
4.  Rearranging part:
    
    重排部分：
    
        For iterator it=m.begin():m.end()-1
            //inner loop to print the character as many times it occurs
            For i=0 : it->second-1 //it->second is the frequency of character it->first
                Print it->first;
            END FOR
        END FOR
        Print sum;

**Example with explanation:**

**带有说明的示例：**

Input:
        XY3BFA9KA2
    
        So,
        After processing :
    
        Map m becomes:
        Char(key)	Int(value)
        'A'	        2
        'B'	        1
        'F'	        1
        'K'	        1
        'X'	        1
        'Y'	        1
    
        Sum: 14
    
        Thus it prints:
        AABFKXY14

**Note:** The [map][maps in C_ STL] is always sorted according to its key. Thus, here we don't need to take any additional care for maintain alphabet order as the [map][maps in C_ STL] itself is always sorted as per its key (character here).

**注意：** [映射][maps in C_ STL]始终根据其键进行排序。 因此，由于[地图][maps in C_ STL]本身始终按其键(此处为字符)进行排序，因此在这里无需额外注意保持字母顺序。

**C++ implementation:**

**C ++实现：**

#include <bits/stdc++.h>
    using namespace std;
    
    void rearrange(string s){
        
    	int sum=0;
    	map<char,int> m;
    	
    	for(int i=0;i<s.length();i++){
        
    		if(s[i]>='0' && s[i]<='9')//digut
    			sum+=s[i]-'0';
    		else //character
    			m[s[i]]++;
    	}
    
    	for(auto it=m.begin();it!=m.end();it++){
        
    		//print the character as many time it occured
    		for(int i=0;i<it->second;i++)
    			cout<<it->first;
    	}
    	cout<<sum<<endl; //printing sum
    }
    
    int main(){
        
    	string s;
    
    	cout<<"Enter input string\n";
    	cin>>s;
    	cout<<"Rearranged string is:\n";
    	rearrange(s);
    
    	return 0;
    }

**Output**

**输出量**

Enter input string
    XY3BFA9KA2
    Rearranged string is:
    AABFKXY14

> 翻译自: [https://www.includehelp.com/icp/rearrange-a-string.aspx][https_www.includehelp.com_icp_rearrange-a-string.aspx]

[maps in C_ STL]: https://www.includehelp.com/stl/map-in-cpp-stl.aspx
[https_www.includehelp.com_icp_rearrange-a-string.aspx]: https://www.includehelp.com/icp/rearrange-a-string.aspx