通过填充元素将给定的字符串拆分为大小为 K 的子字符串

给定一个长度为N的字符串str和一个整数K ，任务是将字符串分成K个大小的组，如果最后一组没有剩余K个字符，则使用字符ch来完成该组。

例子：

Input: str = “Algorithms”, K = 3, ch = “@”
Output: Alg ori thm s@@
Explanation:
The first 3 characters “alg” form the first group.
The next 3 characters “ori” form the second group.
The last 3 characters “thm” form the third group.
For the last group, there is only the character ‘s’ from the string. 
To complete this group, add ‘@’ twice.

Input: str = “Algorithm”, K = 3, ch = “@”
Output: Alg ori thm
Explanation:
Similar to the previous example, 
The first 3 characters “alg” form the first group.
The next 3 characters “ori” form the second group.
The last 3 characters “thm” form the third group.
Since all groups can be completely filled by characters from the string, no need to use ch.

方法：这是一个简单的实现相关问题。请按照以下步骤操作：

1. 将res初始化为空字符串。
2. 开始遍历字符串，当res字符串的大小等于K 时，然后将res字符串放入结果向量中，并再次清空res字符串以进行进一步划分
3. 最后，如果res字符串不为空且 size 不等于 ka，则使用 Extra字符填充它，最后一组。

下面是上述方法的实现。

C++

// C++ code to implement above approach
#include 
using namespace std;
 
// Function to split the string
vector dividestring(string str,
                            int K, char ch)
{
    int N = str.size();
    int j = 0, i = 0;
    vector result;
    string res = "";
    while (j < N) {
        res += str[j];
        if (res.size() == K) {
            result.push_back(res);
            res = "";
        }
        j++;
    }
 
    if (res != "") {
        while (res.size() < K) {
            res += ch;
        }
        result.push_back(res);
    }
    return result;
}
 
// Driver code
int main()
{
 
    string str = "Algorithms";
    int K = 3;
    char ch = '@';
    vector ans
        = dividestring(str, K, ch);
    for (auto i : ans) {
        cout << i << "\n";
    }
    return 0;
}

Java

// Java code to implement above approach
import java.util.ArrayList;
 
class GFG
{
 
  // Function to split the String
  static ArrayList divideString(String str, int K, char ch) {
    int N = str.length();
    int j = 0;
    ArrayList result = new ArrayList();
    String res = "";
    while (j < N) {
      res += str.charAt(j);
      if (res.length() == K) {
        result.add(res);
        res = "";
      }
      j++;
    }
 
    if (res != "") {
      while (res.length() < K) {
        res += ch;
      }
      result.add(res);
    }
    return result;
  }
 
  // Driver code
  public static void main(String args[])
  {
 
    String str = "Algorithms";
    int K = 3;
    char ch = '@';
    ArrayList ans = divideString(str, K, ch);
    for (String i : ans) {
      System.out.println(i);
    }
  }
}
 
// This code is contributed by gfgking.

Python3

# python3 code to implement above approach
 
# Function to split the string
 
 
def dividestring(str, K, ch):
 
    N = len(str)
    j, i = 0, 0
    result = []
    res = ""
    while (j < N):
        res += str[j]
        if (len(res) == K):
            result.append(res)
            res = ""
 
        j += 1
 
    if (res != ""):
        while (len(res) < K):
            res += ch
 
        result.append(res)
 
    return result
 
 
# Driver code
if __name__ == "__main__":
 
    str = "Algorithms"
    K = 3
    ch = '@'
    ans = dividestring(str, K, ch)
    for i in ans:
        print(i)
 
    # This code is contributed by rakeshsahni

C#

// C# code to implement above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to split the string
  static List dividestring(string str, int K,
                                   char ch)
  {
    int N = str.Length;
    int j = 0;
    List result = new List();
    string res = "";
    while (j < N) {
      res += str[j];
      if (res.Length == K) {
        result.Add(res);
        res = "";
      }
      j++;
    }
 
    if (res != "") {
      while (res.Length < K) {
        res += ch;
      }
      result.Add(res);
    }
    return result;
  }
 
  // Driver code
  public static void Main()
  {
 
    string str = "Algorithms";
    int K = 3;
    char ch = '@';
    List ans = new List();
    ans = dividestring(str, K, ch);
    foreach(var i in ans) { Console.WriteLine(i); }
  }
}
 
// This code is contributed by Taranpreet

Javascript

    // JavaScript code for the above approach
     
    // Function to split the string
    function dividestring(str, K, ch)
    {
        let N = str.length;
        let j = 0, i = 0;
        let result = [];
        let res = "";
        while (j < N)
        {
            res += str[j];
            if (res.length == K)
            {
                result.push(res);
                res = "";
            }
            j++;
        }
 
        if (res != "") {
            while (res.length < K) {
                res += ch;
            }
            result.push(res);
        }
        return result;
    }
 
    // Driver code
    let str = "Algorithms";
    let K = 3;
    let ch = '@';
    let ans
        = dividestring(str, K, ch);
    for (let i of ans) {
        document.write(i + '<br>')
    }
 
     // This code is contributed by Potta Lokesh

输出：

Alg
ori
thm
s@@

时间复杂度： O(N)
辅助空间： O(N)