Codility 7번 PermCheck
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].
import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i=0; i<A.length; i++){
map.put(A[i], 1);
}
for(int i=0; i<A.length; i++){
if(!map.containsKey(i+1)){
return 0;
}
}
return 1;
// write your code in Java SE 8
}
}