获得连续的数字而不重复答案

【问题标题】：Getting consecutive numbers without repetition获得连续的数字而不重复
【发布时间】：2025-12-26 10:50:07
【问题描述】：

从一个随机的数字序列{"1", "2", "3", "4", "5"}中，要求确定3个连续数字组出现的次数，表示, 已生成的次数基数 = c ("1", "2", "3", "4", "5") 以下任意组 {"123", "234", " 345" }。

# I undertand that I have generate a sample with 5 numbers
a<-c(sample(1:5,5))
a
#I generated the list, as you can see I didn't fix a seed because I know that in every single sequence I will have differents grupos of 3 consecutive numbers, so I should obtain something like this

b<-c(2,3,4,5,1) #this example gives me just one that it would be {2,3,4}
b
#answer expected
1
#Then, I don't know how to obtain the sequence I have tried with permutations and combinations but I don't get it.

【问题讨论】：

你计算重叠的组吗？如果你有一个序列 1、2、3、4、5，你会数出 123、234、345 吗？还是只有 1 个 123？
{2,4,5} 不连续。它们只是在增加。你想要哪个，连续的还是增加的？
“连续”是否包括环绕？ 451, 512?

标签： r random numbers

【解决方案1】：

这将计算任何出现的三项连续增长（例如 123）。

countsequence <- function(x){
  if (length(x)<3){
    return(0)
  }
  count <- 0
  for(i in 3:length(x)){
    if(x[i-1]==x[i]-1 && x[i-2] == x[i]-2){
      count <- count + 1
    }
  }
  return(count)
} 

countsequence(1:5)
countsequence(c(2, 3, 4, 1, 5))

【讨论】：

如果要连续且不递增，使用if语句如下：if(x[i-1]< x[i] && x[i-2] < x[i-1]){

【解决方案2】：

参数k 用于子序列长度（在您的情况下为 3）：

f <- function(b, k = 3){

  if(k > length(b)) return(0)
  
  check_list <- lapply(k:length(b), function(x) all(diff(b[(x-k+1):x]) == 1)) 
  sum(unlist(check_list))
}

如果您想计算增加或减少的子序列，请将 == 和 1 替换为适当的关系和数字。

【讨论】：