计数排序O（n + k）时间复杂度中的k是多少？

Question

Counting sort worst, best and average time complexity is O(n+k) , where n is number of elements to sort. What is k exactly? I see different definitions: maximum element, difference between max element and min element, and so on.

1. Given array arr1 [1, 3, 5, 9, 12, 7 ] and arr2 [1,2,3,2,1,2,4,1,3,2] what is k for arr1 and arr2 ?
2. Is it true that it is stupid to sort arr1 with counting sort because n < k (element values are from a range which is wider than number of elements to sort?

Answer 1

k is the range of the keys, ie the number of array slots it takes to cover all possible values. Thus in case of numbers, Max-Min+1 . Of course this assumes that you don't waste space by assigning Min the first slot and Max the last.

It is appropriate to use counting sort when k does not exceed a small multiple of n , let nk , as in this case, nk can beat n.log n .

Answer 2

k是数组中的最大可能值，假设您有长度为5的数组，每个数字都是0到9之间的整数，在此示例中，k等于9

Answer 3

First an array of k counts is zeroed. Then the n elements in the array are read, and the elements of the k counts are incremented depending on the values of the n elements. On the output pass of a counting sort, the array of k counts is read, and array of n elements is written. So there are k writes (to zero the counts), n reads, then k reads and n writes for a total of 2n + 2k operations, but big O ignores the constant 2, so the time complexity is O(n + k).