[c#] Codility - PermCheck

<문제>

A non-empty array A consisting of N integers is given.

A permutation is a sequence containing each element from 1 to N once, and only once.

For example, array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

is a permutation, but array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

is not a permutation, because value 2 is missing.

The goal is to check whether array A is a permutation.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A, returns 1 if array A is a permutation and 0 if it is not.

For example, given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3 A[3] = 2

the function should return 1.

Given array A such that:

A[0] = 4 A[1] = 1 A[2] = 3

the function should return 0.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [1..1,000,000,000]

 

<성능>

 

Detected time complexity:O(N) or O(N * log(N))

 

 

<답>

public int Solution(int[] A)
{
  var list = A.ToList();
  list.Sort();

  for (var i = 0; i < list.Count; ++i)
  {
    if (list[0] == 1)
    {
      if (i != list.Count - 1)
      {
        if ((list[i + 1] - list[i] > 1) || (list[i + 1] - list[i] == 0))
        	return 0;
      }
    }
    else
    	return 0;
  }
  return 1;
}

 

<풀이>

- 리스트를 정렬한 다음에, 리스트의 인덱스끼리 비교하였다.