Given an array of ints length 3, return the sum of all the elements.

以下是 Python 代码实现：

def sum3(nums):
    return sum(nums)

或者可以使用以下代码：

def sum3(nums):
    total = 0
    for i in nums:
        total += i
    return total

这两个函数都可以接受一个长度为 3 的整数数组 `nums` 作为输入，返回数组中所有元素之和。

相关问题

Given an array of ints, return the sum of the first 2 elements in the array. If the array length is less than 2, just sum up the elements that exist, returning 0 if the array is length 0.

Here is the Python code to solve the problem:

def sum2(nums):
    if len(nums) == 0:
        return 0
    elif len(nums) < 2:
        return sum(nums)
    else:
        return nums[0] + nums[1]

The `sum2` function takes an array `nums` as input and returns the sum of the first two elements in the array. If the length of the array is less than 2, it sums up all the elements in the array and returns the result. If the array is empty, it returns 0.

You can test the function with different inputs like this:

print(sum2([1, 2, 3])) # Output: 3
print(sum2([1])) # Output: 1
print(sum2([])) # Output: 0

Given an array of ints length 3, figure out which is larger, the first or last element in the array, and set all the other elements to be that value. Return the changed array.

以下是 Python 代码实现：

def max_end3(nums):
    max_num = max(nums[0], nums[-1])
    return [max_num] * 3

这个函数接受一个长度为 3 的整数数组 `nums` 作为输入，首先比较数组的第一个元素和最后一个元素的大小，然后将其他元素设置为较大的那个值，最后返回修改后的数组。

例如，`max_end3([1, 2, 3])` 的返回值为 `[3, 3, 3]`，`max_end3([2, 2, 1])` 的返回值为 `[2, 2, 2]`。