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Codility 4.1 PermCheck

Codility 4.1 PermCheck

作者: 波洛的汽车电子世界 | 来源:发表于2019-08-03 20:00 被阅读0次
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:

def solution(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].

这儿要注意数组的元素是否唯一,可以用set

def solution(A):
    # write your code in Python 3.6
    flag = 0
    s=0
    if len(A) == len(set(A)):
        for i in range(1,len(A)+1):
            s +=i
        if sum(A)==s:
            flag =1
    return flag

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