Question

As mentioned in the title, I want to find the pairs of elements whose difference is K

example k=4 and a[]={7 ,6 23,19,10,11,9,3,15}
output should be :
   7,11
   7,3
   6,10
   19,23
   15,19
   15,11

我阅读了在SO“

In order to find an efficient solution, how much time does it take? Is the time complexity O(nlogn) or O(n)? I tried to do this by a divide and conquer technique, but i m not getting any clue of exit condition...

如果有效的解决办法包括利用两个点对投入阵列和操纵要素进行分类,那么我认为我至少应当使用<代码>O(nlogn)。

是否有任何与数学相关的技术可在O(n)中找到解决办法。感谢任何帮助。

Answer 1

你可以在O(n)以黑板表做。减去O(n)的斜线上的所有数字,然后再一次看<代码>编号[i]+。在O(1),哈希表格中“Yes”或“No”的回报,你需要从所有数字上来,因此总数是O(n)。任何包含O(1)设置和O(1)检查时间的固定结构,都将不采用黑板表。

Answer 2

O(n* Carlo(n)的一个简单解决办法是打平你的阵列,然后穿过你的阵列,履行这一职能:

void find_pairs(int n, int array[], int k)
{
  int first = 0;
  int second = 0;
  while (second < n)
  {
    while (array[second] < array[first]+k)
      second++;
    if (array[second] == array[first]+k)
      printf("%d, %d
", array[first], array[second]);
    first++;
  }
}

这种解决办法没有使用与解决办法不同的额外空间。

Answer 3

One thing may be done using indexing in O(n)

Take a boolean array arr indexed by the input list.
For each integer i is in the input list then set arr[i] = true
Traverse the entire arr from the lowest integer to the highest integer as follows:
- whenever you find a true at ith index, note down this index.
- see if there arr[i+k] is true. If yes then i and i+k numbers are the required pair
- else continue with the next integer i+1

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