문제 : Counting Sort 1
사용언어 : C#
문제 내용
Comparison Sorting
Quicksort usually has a running time of n x log(n), but is there an algorithm that can sort even faster? In general, this is not possible. Most sorting algorithms are comparison sorts, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beat n x log(n) (worst-case) running time, since n x log(n) represents the minimum number of comparisons needed to know where to place each element. For more details, you can see these notes (PDF).
Alternative Sorting
Another sorting method, the counting sort, does not require comparison. Instead, you create an integer array whose index range covers the entire range of values in your array to sort. Each time a value occurs in the original array, you increment the counter at that index. At the end, run through your counting array, printing the value of each non-zero valued index that number of times.
Example
arr = [1,1,3,2,1]
All of the values are in the range [0 ... 3], so create an array of zeros, result = [0,0,0,0]. The results of each iteration follow:
i arr[i] result
0 1 [0, 1, 0, 0]
1 1 [0, 2, 0, 0]
2 3 [0, 2, 0, 1]
3 2 [0, 2, 1, 1]
4 1 [0, 3, 1, 1]
The frequency array is [0,3,1,1]. These values can be used to create the sorted array as well: sorted = [1,1,1,2,3].
Note
For this exercise, always return a frequency array with 100 elements. The example above shows only the first 4 elements, the remainder being zeros.
Challenge
Given a list of integers, count and return the number of times each value appears as an array of integers.
Function Description
Complete the countingSort function in the editor below.
countingSort has the following parameter(s):
- arr[n]: an array of integers
Returns
- int[100]: a frequency array
Input Format
The first line contains an integer n, the number of items in arr.
Each of the next lines contains an integer arr[i] where 0 <= i < n.
Constraints
100 <= n <= 10^6
0 <= arr[i] < 100
Sample Input
100
63 25 73 1 98 73 56 84 86 57 16 83 8 25 81 56 9 53 98 67 99 12 83 89 80 91 39 86 76 85 74 39 25 90 59 10 94 32 44 3 89 30 27 79 46 96 27 32 18 21 92 69 81 40 40 34 68 78 24 87 42 69 23 41 78 22 6 90 99 89 50 30 20 1 43 3 70 95 33 46 44 9 69 48 33 60 65 16 82 67 61 32 21 79 75 75 13 87 70 33
Sample Output
0 2 0 2 0 0 1 0 1 2 1 0 1 1 0 0 2 0 1 0 1 2 1 1 1 3 0 2 0 0 2 0 3 3 1 0 0 0 0 2 2 1 1 1 2 0 2 0 1 0 1 0 0 1 0 0 2 1 0 1 1 1 0 1 0 1 0 2 1 3 2 0 0 2 1 2 1 0 2 2 1 2 1 2 1 1 2 2 0 3 2 1 1 0 1 1 1 0 2 2
Explanation
Each of the resulting values result[i] represents the number of times i appeared in arr.
내 풀이
받아온 list에서 각 요소의 빈도수를 다시 list로 만드는 것이 문제였다.
처음에는 문제를 잘못 봐서 처음 배열 길이만큼의 배열을 만들었는데
알고 보니까 배열의 크기는 100으로 만들라는 거였다.
크기가 100인 list하나를 만들고 인자값으로 받아온 arr의 요소를 foreach로 하나씩 돌면서
자신의 자리에 +1을 해서 만들었다.
private static List<int> nums;
public static List<int> countingSort(List<int> arr)
{
nums = new List<int>(new int[100]);
foreach (int i in arr)
{
nums[i] += 1;
}
return nums;
}