给定数字N。任务是找到Golomb序列的前N个项。 Golomb序列是一个非递减整数序列，其中第n个项等于n在序列中出现的次数。
Input: N = 11
Output: 1 2 2 3 3 4 4 4 5 5 5
Explanation:
The first term is 1. 1 appears only once.
The second term is 2. 2 appears two times.
The third term is 2. 3 appears two times.
The fourth term is 3. 4 appears three times.
the fifth term is 3. 5 appears three times.
Input: N = 6
Output: 1 2 2 3 3 4
方法：
- 用[1]初始化数组arr [] ，因为Golomb序列以1开头。
- 将最后一个索引的值存储在映射M中。
- 对于Golomb序列，直到N ：
- 初始化cnt = 0并映射。
- 如果cnt不等于0，则Golomb序列中的当前元素是该序列中的前一个元素，并降低cnt。
- 否则，Golomb序列中的当前元素等于1 +序列中的前一个元素，并且cnt被更新为Golomb序列中当前元素处map的值，并降低了cnt。
- 使用Golomb序列中的当前值映射当前索引。
- 打印存储在数组arr []中的Golomb序列的所有元素。
下面是上述方法的实现：
C++
// C++ program to find the first
// N terms of Golomb Sequence
#include "bits/stdc++.h"
#define MAX 100001
using namespace std;
// Function to print the Golomb
// Sequence
void printGolombSequence(int N)
{
// Initialise the array
int arr[MAX];
// Initialise the cnt to 0
int cnt = 0;
// First and second element
// of Golomb Sequence is 0, 1
arr[0] = 0;
arr[1] = 1;
// Map to store the count of
// current element in Golomb
// Sequence
map M;
// Store the count of 2
M[2] = 2;
// Iterate over 2 to N
for (int i = 2; i <= N; i++) {
// If cnt is equals to 0
// then we have new number
// for Golomb Sequence
// which is 1 + previous
// element
if (cnt == 0) {
arr[i] = 1 + arr[i - 1];
cnt = M[arr[i]];
cnt--;
}
// Else the current element
// is the previous element
// in this Sequence
else {
arr[i] = arr[i - 1];
cnt--;
}
// Map the current index to
// current value in arr[]
M[i] = arr[i];
}
// Print the Golomb Sequence
for (int i = 1; i <= N; i++) {
cout << arr[i] << ' ';
}
}
// Driver Code
int main()
{
int N = 11;
printGolombSequence(N);
return 0;
} Java
// Java program to find the first
// N terms of Golomb Sequence
import java.util.*;
class GFG{
static int MAX = 1000;
// Function to print the Golomb
// Sequence
static void printGolombSequence(int N)
{
// Initialise the array
int []arr = new int[MAX];
for(int i = 0; i < MAX; i++)
arr[i] = 0;
// Initialise the cnt to 0
int cnt = 0;
// First and second element
// of Golomb Sequence is 0, 1
arr[0] = 0;
arr[1] = 1;
// Map to store the count of
// current element in Golomb
// Sequence
Map M=new HashMap();
// Store the count of 2
M.put(2,2);
// Iterate over 2 to N
for (int i = 2; i <= N; i++) {
// If cnt is equals to 0
// then we have new number
// for Golomb Sequence 1 2 2 3 3 4 4 4 5 5 5
// which is 1 + previous
// element
if (cnt == 0) {
arr[i] = 1 + arr[i - 1];
cnt = M.get(arr[i]);
cnt--;
}
// Else the current element
// is the previous element
// in this Sequence
else {
arr[i] = arr[i - 1];
cnt--;
}
// Map the current index to
// current value in arr[]
M.put(i, arr[i]);
}
// Print the Golomb Sequence
for (int i = 1; i <= N; i++)
{
System.out.print(arr[i]+" ");
}
}
// Driver Code
public static void main(String args[])
{
int N = 11;
printGolombSequence(N);
}
}
// This code is contributed by Surendra_Gangwar Python3
# Python3 program to find the first
# N terms of Golomb Sequence
MAX = 100001
# Function to print the Golomb
# Sequence
def printGolombSequence(N):
# Initialise the array
arr = [0] * MAX
# Initialise the cnt to 0
cnt = 0
# First and second element
# of Golomb Sequence is 0, 1
arr[0] = 0
arr[1] = 1
# Map to store the count of
# current element in Golomb
# Sequence
M = dict()
# Store the count of 2
M[2] = 2
# Iterate over 2 to N
for i in range(2, N + 1):
# If cnt is equals to 0
# then we have new number
# for Golomb Sequence
# which is 1 + previous
# element
if (cnt == 0):
arr[i] = 1 + arr[i - 1]
cnt = M[arr[i]]
cnt -= 1
# Else the current element
# is the previous element
# in this Sequence
else:
arr[i] = arr[i - 1]
cnt -= 1
# Map the current index to
# current value in arr[]
M[i] = arr[i]
# Print the Golomb Sequence
for i in range(1, N + 1):
print(arr[i], end=" ")
# Driver Code
N = 11
printGolombSequence(N)
# This code is contributed by mohit kumar 29C#
// C# program to find the first
// N terms of Golomb Sequence
using System;
using System.Collections.Generic;
class GFG{
static int MAX = 1000;
// Function to print the Golomb
// Sequence
static void printGolombSequence(int N)
{
// Initialise the array
int[] arr = new int[MAX];
for(int i = 0; i < MAX; i++)
arr[i] = 0;
// Initialise the cnt to 0
int cnt = 0;
// First and second element
// of Golomb Sequence is 0, 1
arr[0] = 0;
arr[1] = 1;
// Map to store the count of
// current element in Golomb
// Sequence
Dictionary M = new Dictionary();
// Store the count of 2
M.Add(2, 2);
// Iterate over 2 to N
for(int i = 2; i <= N; i++)
{
// If cnt is equals to 0
// then we have new number
// for Golomb Sequence 1 2 2 3 3 4 4 4 5 5 5
// which is 1 + previous
// element
if (cnt == 0)
{
arr[i] = 1 + arr[i - 1];
cnt = M[arr[i]];
cnt--;
}
// Else the current element
// is the previous element
// in this Sequence
else
{
arr[i] = arr[i - 1];
cnt--;
}
// Map the current index to
// current value in arr[]
if(M.ContainsKey(i))
{
M[i] = arr[i];
}
else
{
M.Add(i, arr[i]);
}
}
// Print the Golomb Sequence
for(int i = 1; i <= N; i++)
{
Console.Write(arr[i] + " ");
}
}
// Driver Code
static void Main()
{
int N = 11;
printGolombSequence(N);
}
}
// This code is contributed by divyeshrabadiya07 输出：
1 2 2 3 3 4 4 4 5 5 5时间复杂度： O(N)
辅助空间： O(N)