打印从前n个自然数开始的所有长度为k的递增序列

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📜 打印从前n个自然数开始的所有长度为k的递增序列

📅 最后修改于: 2021-05-07 00:46:37 🧑 作者: Mango

给定两个正整数n和k，打印所有长度为k的递增序列，以使每个序列中的元素都来自前n个自然数。

例子：

Input: k = 2, n = 3
Output: 1 2
        1 3
        2 3 

Input: k = 5, n = 5
Output: 1 2 3 4 5

Input: k = 3, n = 5
Output: 1 2 3
        1 2 4
        1 2 5
        1 3 4
        1 3 5
        1 4 5
        2 3 4
        2 3 5
        2 4 5
        3 4 5

强烈建议您最小化浏览器，然后自己尝试。
这是一个很好的递归问题。这个想法是创建一个长度为k的数组。数组存储当前序列。对于数组中的每个位置，我们检查前一个元素，然后一个接一个地放置所有大于前一个元素的元素。如果没有上一个元素(第一个位置)，我们将所有数字从1到n。

下面是上述想法的实现：

C++

// C++ program to  print all increasing sequences of
// length 'k' such that the elements in every sequence
// are from first 'n' natural numbers.
#include
using namespace std;
 
// A utility function to print contents of arr[0..k-1]
void printArr(int arr[], int k)
{
    for (int i=0; i




Java

// Java program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n'
// natural numbers.
 
class GFG {
     
    // A utility function to print
    // contents of arr[0..k-1]
    static void printArr(int[] arr, int k)
    {
        for (int i = 0; i < k; i++)
            System.out.print(arr[i] + " ");
        System.out.print("\n");
    }
     
    // A recursive function to print
    // all increasing sequences
    // of first n natural numbers.
    // Every sequence should be
    // length k. The array arr[] is
    // used to store current sequence
    static void printSeqUtil(int n, int k,
                             int len, int[] arr)
    {
         
        // If length of current increasing
        // sequence becomes k, print it
        if (len == k)
        {
            printArr(arr, k);
            return;
        }
     
        // Decide the starting number
        // to put at current position:
        // If length is 0, then there
        // are no previous elements
        // in arr[]. So start putting
        // new numbers with 1.
        // If length is not 0,
        // then start from value of
        // previous element plus 1.
        int i = (len == 0) ? 1 : arr[len - 1] + 1;
     
        // Increase length
        len++;
     
        // Put all numbers (which are
        // greater than the previous
        // element) at new position.
        while (i <= n)
        {
            arr[len - 1] = i;
            printSeqUtil(n, k, len, arr);
            i++;
        }
     
        // This is important. The
        // variable 'len' is shared among
        // all function calls in recursion
        // tree. Its value must be
        // brought back before next
        // iteration of while loop
        len--;
    }
     
    // This function prints all
    // increasing sequences of
    // first n natural numbers.
    // The length of every sequence
    // must be k. This function
    // mainly uses printSeqUtil()
    static void printSeq(int n, int k)
    {
         
        // An array to store
        // individual sequences
        int[] arr = new int[k];
         
        // Initial length of
        // current sequence
        int len = 0;
        printSeqUtil(n, k, len, arr);
    }
 
    // Driver Code
    static public void main (String[] args)
    {
        int k = 3, n = 7;
        printSeq(n, k);
    }
}
 
// This code is contributed by Smitha.



Python3

# Python3 program to print all
# increasing sequences of length
# 'k' such that the elements in
# every sequence are from first
# 'n' natural numbers.
 
# A utility function to
# print contents of arr[0..k-1]
def printArr(arr, k):
    for i in range(k):
        print(arr[i], end = " ");
    print();
 
# A recursive function to print
# all increasing sequences of
# first n natural numbers. Every
# sequence should be length k.
# The array arr[] is used to
# store current sequence.
def printSeqUtil(n, k,len1, arr):
     
    # If length of current
    # increasing sequence
    # becomes k, print it
    if (len1 == k):
        printArr(arr, k);
        return;
 
    # Decide the starting number
    # to put at current position:
    # If length is 0, then there
    # are no previous elements
    # in arr[]. So start putting
    # new numbers with 1. If length
    # is not 0, then start from value
    # of previous element plus 1.
    i = 1 if(len1 == 0) else (arr[len1 - 1] + 1);
 
    # Increase length
    len1 += 1;
 
    # Put all numbers (which are greater
    # than the previous element) at
    # new position.
    while (i <= n):
        arr[len1 - 1] = i;
        printSeqUtil(n, k, len1, arr);
        i += 1;
 
    # This is important. The variable
    # 'len' is shared among all function
    # calls in recursion tree. Its value
    # must be brought back before next
    # iteration of while loop
    len1 -= 1;
 
# This function prints all increasing
# sequences of first n natural numbers.
# The length of every sequence must be
# k. This function mainly uses printSeqUtil()
def printSeq(n, k):
        arr = [0] * k; # An array to store
                       # individual sequences
        len1 = 0; # Initial length of
                  # current sequence
        printSeqUtil(n, k, len1, arr);
 
# Driver Code
k = 3;
n = 7;
printSeq(n, k);
 
# This code is contributed by mits



C#

// C# program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n'
// natural numbers.
using System;
 
class GFG {
     
    // A utility function to print
    // contents of arr[0..k-1]
    static void printArr(int[] arr, int k)
    {
        for (int i = 0; i < k; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
    }
     
    // A recursive function to print
    // all increasing sequences
    // of first n natural numbers.
    // Every sequence should be
    // length k. The array arr[] is
    // used to store current sequence
    static void printSeqUtil(int n, int k,
                             int len, int[] arr)
    {
         
        // If length of current increasing
        // sequence becomes k, print it
        if (len == k)
        {
            printArr(arr, k);
            return;
        }
     
        // Decide the starting number
        // to put at current position:
        // If length is 0, then there
        // are no previous elements
        // in arr[]. So start putting
        // new numbers with 1.
        // If length is not 0,
        // then start from value of
        // previous element plus 1.
        int i = (len == 0) ? 1 : arr[len - 1] + 1;
     
        // Increase length
        len++;
     
        // Put all numbers (which are
        // greater than the previous
        // element) at new position.
        while (i <= n)
        {
            arr[len - 1] = i;
            printSeqUtil(n, k, len, arr);
            i++;
        }
     
        // This is important. The
        // variable 'len' is shared among
        // all function calls in recursion
        // tree. Its value must be
        // brought back before next
        // iteration of while loop
        len--;
    }
     
    // This function prints all
    // increasing sequences of
    // first n natural numbers.
    // The length of every sequence
    // must be k. This function
    // mainly uses printSeqUtil()
    static void printSeq(int n, int k)
    {
         
        // An array to store
        // individual sequences
        int[] arr = new int[k];
         
        // Initial length of
        // current sequence
        int len = 0;
        printSeqUtil(n, k, len, arr);
    }
 
    // Driver Code
    static public void Main ()
    {
        int k = 3, n = 7;
        printSeq(n, k);
    }
}
 
// This code is contributed by Ajit.



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