具有不同后续字符的字符串的连续子段
给定长度为 L 的字符串和一个整数 N ，任务是形成包含不同后续字符的字符串的总共 (L / N) 个连续子段。 注意:整数 N 将是字符串长度的一个因子，即 L 。
示例:
输入:str = " geesforgeeksgfg "，N = 4 输出: gek for gek sgf “geesforgeeksgfg”的长度为 16，因此会有 4 个子段。 第一个子段将包含字符 g、e、e 和 k，但是字母‘e’是重复的， 所以我们丢弃一个‘e’。因此，最终的子片段将是“gek”。 同样，其他三个子段将为“稳定”、“gek”和“sgf”。 每个亚片段应具有随后的不同特征。
输入:str = " aabdefgf "，N = 3 输出: ab dek fg
方法:每次在新的子段上开始迭代时，都会创建一个数组。这些元素按顺序存储在该数组中。如果数组中已经存在任何元素，则不会将其推入数组。 以下是上述办法的实施情况:
卡片打印处理机（Card Print Processor 的缩写）
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function that prints the segments
void sub_segments(string str, int n)
{
int l = str.length();
for (int x = 0; x < l; x += n)
{
string newlist = str.substr(x, n);
// New array for every iteration
list<char> arr;
// Iterator for new array
list<char>::iterator it;
for(auto y:newlist)
{
it = find(arr.begin(), arr.end(), y);
// Check if iterator points to end or not
if(it == arr.end())
arr.push_back(y);
}
for(auto y:arr)
cout << y;
cout << endl;
}
}
// Driver code
int main()
{
string str = "geeksforgeeksgfg";
int n = 4;
sub_segments(str, n);
}
// This code is contributed by Sanjit_Prasad
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that prints the segments
static void sub_segments(String str, int n)
{
int l = str.length();
for (int x = 0; x < l; x += n)
{
String newlist = str.substring(x, x + n);
// New array for every iteration
List<Character> arr = new ArrayList<Character>();
for (char y : newlist.toCharArray())
{
// Check if the character is in the array
if (!arr.contains(y))
arr.add(y);
}
for (char y : arr)
System.out.print(y);
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeksgfg";
int n = 4;
sub_segments(str, n);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function that prints the segments
def sub_segments (string, n):
l = len (string)
for x in range (0, l, n):
newlist = string[x : x + n]
# New array for every iteration
arr = []
for y in newlist:
# Check if the character is in the array
if y not in arr:
arr.append (y)
print (''.join (arr))
# Driver code
string = "geeksforgeeksgfg"
n = 4
sub_segments (string, n)
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that prints the segments
static void sub_segments(String str, int n)
{
int l = str.Length;
for (int x = 0; x < l; x += n)
{
String newlist = str.Substring(x, n);
// New array for every iteration
List<char> arr = new List<char>();
foreach (char y in newlist.ToCharArray())
{
// Check if the character is in the array
if (!arr.Contains(y))
arr.Add(y);
}
foreach (char y in arr)
Console.Write(y);
Console.WriteLine();
}
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeksgfg";
int n = 4;
sub_segments(str, n);
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// Javascript implementation of the approach
// Function that prints the segments
function sub_segments(str, n) {
let l = str.length;
for (let x = 0; x < l; x += n) {
let newlist = str.substr(x, n);
// New array for every iteration
let arr = [];
for (let y of newlist) {
// Check if the character is in the array
if (!arr.includes(y))
arr.push(y);
}
for (let y of arr)
document.write(y);
document.write("<br>");
}
}
// Driver code
let str = "geeksforgeeksgfg";
let n = 4;
sub_segments(str, n);
// This code is contributed by gfgking
</script>
Output:
gek
sfor
gek
sgf
版权属于：月萌API www.moonapi.com，转载请注明出处
