将两个字符串合并成给定大小的块
给定两个字符串'a' 和 'b' 以及一个数字 k，我们的目标是将这些字符串合并为一个字符串s ，使其在最终字符串中交替包含来自字符串的 k 个字符组。
例子：
Input: a = "durability",
b = "essence
k = 3
Output: duressabienclitey
Input: a = "determination"
b = "stance"
k = 3
Output: detstaermnceination在第二个例子中，我们交替地从决心和立场中选择三个字符为一组来制作字符串。但是当完全使用字符串姿势时，我们将剩余的决心字符串按原样添加。
这个问题的方法是交替添加两个字符串的字符组，直到一个字符串完成。然后我们只需将剩余字符串的剩余部分原样添加到答案中。
C++
// C++ program to merge n number of strings
#include
using namespace std;
// Function performing the calculations
void solve(string a, string b, int k)
{
string s = "";
// Length of string a
int la = a.length();
// Length f string b
int lb = b.length();
int l = la + lb;
// Pointers for string
// a and string b
int indexa = 0, indexb = 0;
while (l > 0) {
// pa and pb denote the number
// of characters of both
// a and b extracted
int pa = 0, pb = 0;
// If entire substring of length
// k can be extracted
if (la - indexa >= k) {
// Please refer below link for
// details of library function
// https://www.geeksforgeeks.org/stdsubstr-in-ccpp/
s = s + a.substr(indexa, k);
indexa = indexa + k;
pa = k;
}
// If the remaining string is
// of length less than k
else if (la - indexa < k && la - indexa > 0) {
s = s + a.substr(indexa, la - indexa);
pa = la - indexa;
indexa = la;
}
// If the string has been
// traversed
else if (indexa >= la)
pa = 0;
// If entire substring of
// length k can be extracted
if (lb - indexb >= k) {
s = s + b.substr(indexb, k);
pb = k;
indexb = indexb + k;
}
// If the remaining string is of
// length less than k
else if (lb - indexb < k && lb - indexb > 0) {
s = s + b.substr(indexb, lb - indexb);
pb = lb - indexb;
indexb = lb;
}
// If the string has been
// traversed
else if (indexb >= lb)
pb = 0;
l = l - pa - pb;
}
cout << s;
}
// Driver function
int main()
{
string a = "determination", b = "stance";
int k = 3;
solve(a, b, k);
return 0;
} Java
// Java program to merge
// n number of strings
import java.io.*;
class msc {
// Function performing calculations
public static void solve(String a,
String b, int k)
{
String s = "";
// Length of string a
int la = a.length();
// Length f string b
int lb = b.length();
int l = la + lb;
// Pointers for string a and string b
int indexa = 0, indexb = 0;
while (l > 0) {
// pa and pb denote the number
// of characters of both
// a and b extracted
int pa = 0, pb = 0;
// If entire substring of
// length k can be extracted
if (la - indexa >= k) {
s = s + a.substring(indexa, indexa + k);
indexa = indexa + k;
pa = k;
}
// If the remaining string is
// of length less than k
else if (la - indexa < k && la - indexa > 0) {
s = s + a.substring(indexa, la);
pa = la - indexa;
indexa = la;
}
// If the string has been
// traversed
else if (indexa >= la)
pa = 0;
// If entire substring of
// length k can be extracted
if (lb - indexb >= k) {
s = s + b.substring(indexb, indexb + k);
pb = k;
indexb = indexb + k;
}
// If the remaining string
// is of length less than k
else if (lb - indexb < k && lb - indexb > 0) {
s = s + b.substring(indexb, lb);
pb = lb - indexb;
indexb = lb;
}
// If the string has been
// traversed
else if (indexb >= lb)
pb = 0;
l = l - pa - pb;
}
System.out.println(s);
}
// Driver function
public static void main(String args[])
throws IOException
{
String a = "determination", b = "stance";
int k = 3;
solve(a, b, k);
}
}Python3
# Python3 program to merge n number of strings
# Function performing the calculations
def solve(a, b, k):
s = ""
# Length of string a
la = len(a)
# Length f string b
lb = len(b)
l = la + lb
# Pointers for string
# a and string b
indexa = indexb = 0
while l:
# pa and pb denote the number
# of characters of both
# a and b extracted
pa = pb = 0
# If entire substring of length
# k can be extracted
if la - indexa >= k:
s = s + a[indexa : indexa + k]
indexa = indexa + k
pa = k
# If the remaining string is
# of length less than k
elif la - indexa < k and la - indexa:
s = s + a[indexa : la]
pa = la - indexa
indexa = la
# If the string has been
# traversed
elif indexa >= la:
pa = 0
# If entire substring of
# length k can be extracted
if lb - indexb >= k:
s = s + b[indexb : indexb+k]
pb = k
indexb = indexb + k
# If the remaining string is of
# length less than k
elif lb - indexb < k and lb - indexb:
s = s + b[indexb : lb]
pb = lb - indexb
indexb = lb
# If the string has been
# traversed
elif indexb >= lb:
pb = 0
l = l - pa - pb
print(s)
# Driver function
a = "determination"; b = "stance"
k = 3
solve(a, b, k)
# This code is contributed by Ansu KumariC#
// C# program to merge
// n number of strings
using System;
class GFG
{
// Function performing
// the calculations
static void solve(string a,
string b,
int k)
{
string s = "";
// Length of string a
int la = a.Length;
// Length f string b
int lb = b.Length;
int l = la + lb;
// Pointers for string
// a and string b
int indexa = 0, indexb = 0;
while (l > 0)
{
// pa and pb denote the
// number of characters
// a and b extracted
int pa = 0, pb = 0;
// If entire substring
// of length k can be
// extracted
if (la - indexa >= k)
{
s = s + a.Substring(indexa, k);
indexa = indexa + k;
pa = k;
}
// If the remaining string
// is of length less than k
else if (la - indexa < k &&
la - indexa > 0)
{
s = s + a.Substring(indexa,
la - indexa);
pa = la - indexa;
indexa = la;
}
// If the string has
// been traversed
else if (indexa >= la)
pa = 0;
// If entire substring
// of length k can be
// extracted
if (lb - indexb >= k)
{
s = s + b.Substring(indexb, k);
pb = k;
indexb = indexb + k;
}
// If the remaining string
// is of length less than k
else if (lb - indexb < k &&
lb - indexb > 0)
{
s = s + b.Substring(indexb,
lb - indexb);
pb = lb - indexb;
indexb = lb;
}
// If the string has
// been traversed
else if (indexb >= lb)
pb = 0;
l = l - pa - pb;
}
Console.Write(s);
}
// Driver Code
static void Main()
{
string a = "determination",
b = "stance";
int k = 3;
solve(a, b, k);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)输出 ：
detstaermnceination