Check whether array A is a permutation.
Task description
A non-empty zero-indexed 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.
The goal is to check whether array A is a permutation.
Write a function:
int solution(int A[], int N);
that, given a zero-indexed 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.
Assume that:
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].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
Copyright 2009�C2014 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution C:42
#define FLAG(N) (((N)>>30)&0x3)
#define VALUE(N) ((N)&(0xffffffff>>2))
#define SET_FLAG(N, flag) ((N)|((flag)<<30))
#define NOT_VISITED 0
#define VISITED 1
int solution(int A[], int N)
{
int notFound = N;
char* reason = "good";
int i = 0;
for( i = 0; i < N; i++ )
{
int value = VALUE(A[i]);
if(value > N)
{
// 数字超过N,不可能是全排列
reason = "overflow";
break;
}
int next = value - 1;
if( FLAG(A[next]) == NOT_VISITED )
{
A[next] = SET_FLAG( A[next], VISITED);
notFound --;
}
else
{
// 出现重复的元素, 不可能是全排列
reason = "duplicate";
break;
}
}
//printf("reason:%s", reason);
return (notFound == 0);
}
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