Find the smallest integer in the array 使用python语法

可以使用Python中的min函数来找到数组中的最小值，示例如下：

arr = [3, 5, 2, 8, 1]
smallest = min(arr)
print(smallest) # 输出1

其中，arr是一个包含整数的数组，min函数返回数组中的最小值，将其赋值给变量smallest，并打印输出。

相关问题

(Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

Here's a Python program that implements the algorithm described in the problem statement:

import math

def smallest_perfect_square_factor(m):
    factors = []
    # Find all smallest factors of m
    while m % 2 == 0:
        factors.append(2)
        m = m // 2
    for i in range(3, int(math.sqrt(m))+1, 2):
        while m % i == 0:
            factors.append(i)
            m = m // i
    if m > 2:
        factors.append(m)
    # Compute n as the product of factors that appear an odd number of times
    odd_factors = [x for x in factors if factors.count(x) % 2 == 1]
    n = 1
    for f in odd_factors:
        n *= f
    return n

m = int(input("Enter an integer m: "))
n = smallest_perfect_square_factor(m)
print(f"The smallest integer n such that m * n is a perfect square is {n}.")

The program first computes all the smallest factors of `m` and stores them in an array list called `factors`. Then, it identifies the factors that appear an odd number of times in the list, and multiplies them together to obtain `n`. Finally, the program prints out the value of `n`.

编写一段python代码，解决以下问题：Given an unsorted integer array nums, return the smallest missing positive integer. You must implement an algorithm that runs in O(n) time and uses constant extra space.

这是给出的解决方案：

def findSmallestMissingPositive(nums):
    # 将所有小于等于 0 的数替换为 len(nums) + 1
    for i in range(len(nums)):
        if nums[i] <= 0:
            nums[i] = len(nums) + 1

    # 将所有在范围内的数替换为负数
    for i in range(len(nums)):
        num = abs(nums[i])
        if num <= len(nums):
            nums[num - 1] = -abs(nums[num - 1])

    # 找到第一个正数，它就是缺失的最小正数
    for i in range(len(nums)):
        if nums[i] > 0:
            return i + 1

    # 如果数组中的所有数都是负数，则返回 len(nums) + 1
    return len(nums) + 1

这个算法在时间复杂度上做到了 O(n)，因为它只遍历了数组 nums 一次。它还使用了常数的额外空间，因为它只使用了一些变量来存储数据。