如何检查列表是否是python中的连续数字？

Question

How to write python code that let the computer know if the list is a right sequence and the position doesn't matter, it will return true, otherwise it return false. below are some of my example, I really don't know how to start

b=[1,2,3,4,5] #return true
b=[1,2,2,1,3] # return false
b=[2,3,1,5,4] #return true
b=[2,4,6,4,3] # return false 

Answer 1

sort function is O(nlogn), we can use for loop which is O(n):

def check_seq(in_list):
    now_ele = set()
    min_ele = max_ele = in_list[0]

    for i in in_list:
        if i in now_ele:
            return False
        min_ele = min(i, min_ele)
        max_ele = max(i, max_ele)
        now_ele.add(i)
    
    if max_ele-min_ele+1 == len(in_list):
        return True
    return False

Answer 2

根据最小值和最大值创建一组和一个进行比较：

isRightSequence = set(range(min(b), max(b)+1)) == set(b)

Answer 3

This question is quite simple and can be solved a few ways.

1. The conditional approach - if there is a number that is bigger than the length of the list, it automatically cannot be a sequence because there can only be numbers from 1-n where n is the size of the list. Also, you have to check if there are any duplicates in the list as this cannot be possible either. If none of these conditions occur, it should return true

2. Using dictionary - go through the entire list and add it as a key to a dictionary. Afterwards, simply loop through numbers 1-n where n is the length of the list and check if they are keys in the dictionary, if one of them isn't, return false. If all of them are, return true.

Both of these are quite simply approaches and you should be able to implement them yourselves. However, this is one implementation for both.

1.

def solve(list1):
    seen = {}
    for i in list1:
        if i > len(list1):
            return False
        if i in seen:
            return False
        seen[i] = True
    return False

def solve(list1):
    seen = {}
    for i in list1:
        seen[i] = True
    for i in range (1, len(list1)+1):
        if i not in seen:
            return False
    return True

Answer 4

This solution needs O(n) runtime and O(n) space

def is_consecutive(l: list[int]):
    if not l: 
        return False
    low = min(l)
    high = max(l)
    # Bounds Check
    if high - low != len(l) - 1:
        return False
    
    # Test all indices exist
    test_vec = [False] * len(l) # O(n)
    for i in range(len(l)): 
        test_vec[l[i] - low] = True
    return all(test_vec)

assert is_consecutive(range(10))
assert is_consecutive([-1, 1,0])
assert not is_consecutive([1,1])
assert not is_consecutive([1,2,4,6,5])