蓝桥杯上岸每日N题第一期(二)！！！

**蓝桥杯上岸每日N题第一期(二)****前言**

蓝桥杯是中国的一项非常著名的编程竞赛活动，由蓝桥网主办。它旨在培养学生的编程能力、逻辑思维和问题解决能力等方面。蓝桥杯上岸每日N题第一期(二)是本次竞赛的第二部分。

**第一题：**

**题目描述**

给定一个整数数组，要求找出其中最长的连续子序列，其元素之和为零。

**示例**

输入：[-1,0,1,2, -1, -4]

输出：3（因为 [-1,-1,2] 是最长的连续子序列，其元素之和为零）

**代码实现**

def longest_zero_sum_subarray(nums):
 """
 Finds the longest continuous subarray with a sum of zero.

 Args:
 nums (list): A list of integers.

 Returns:
 int: The length of the longest continuous subarray with a sum of zero.
 """
 max_length =0 current_sum =0 # Initialize two pointers, one at the start and one at the end of the array left =0 right =0 while right < len(nums):
 # Add the element at the right pointer to the current sum current_sum += nums[right]

 # If the current sum is zero, update the max length and move both pointers if current_sum ==0:
 max_length = max(max_length, right - left +1)
 left = right # If the current sum is negative, reset it to zero and move the left pointer elif current_sum < 0:
 current_sum =0 left = right # Move the right pointer right +=1 return max_length

**注释**

* 我们使用了两个指针，分别从左到右和从右到左扫描数组。
* 当我们遇到一个元素时，我们将其添加到当前和中。如果当前和为零，我们更新最大长度并移动两个指针。
* 如果当前和为负数，我们重置它为零并移动左指针。

**第二题：**

**题目描述**

给定一个整数数组，要求找出其中最长的连续子序列，其元素之差为1。

**示例**

输入：[1,2,3,4,5]

输出：5（因为 [1,2,3,4,5] 是最长的连续子序列，其元素之差为1）

**代码实现**

def longest_one_diff_subarray(nums):
 """
 Finds the longest continuous subarray with a difference of one.

 Args:
 nums (list): A list of integers.

 Returns:
 int: The length of the longest continuous subarray with a difference of one.
 """
 max_length =1 current_length =1 # Iterate over the array from left to right for i in range(1, len(nums)):
 # If the difference between the current and previous elements is one,
 # increment the current length if nums[i] - nums[i -1] ==1:
 current_length +=1 # Otherwise, reset the current length to one else:
 current_length =1 # Update the max length if necessary max_length = max(max_length, current_length)

 return max_length

**注释**

* 我们使用了一个变量 `current_length` 来跟踪当前连续子序列的长度。
* 当我们遇到一个元素时，我们检查它与前一个元素之间的差值。如果差值为1，我们增加 `current_length`。否则，我们重置 `current_length` 为1。
* 我们更新最大长度 `max_length` 如果当前长度大于它。

**第三题：**

**题目描述**

给定一个整数数组，要求找出其中最长的连续子序列，其元素之和为奇数。

**示例**

输入：[1,2,3,4,5]

输出：5（因为 [1,2,3,4,5] 是最长的连续子序列，其元素之和为奇数）

**代码实现**

def longest_odd_sum_subarray(nums):
 """
 Finds the longest continuous subarray with an odd sum.

 Args:
 nums (list): A list of integers.

 Returns:
 int: The length of the longest continuous subarray with an odd sum.
 """
 max_length =0 current_sum =0 # Initialize two pointers, one at the start and one at the end of the array left =0 right =0 while right < len(nums):
 # Add the element at the right pointer to the current sum current_sum += nums[right]

 # If the current sum is odd, update the max length and move both pointers if current_sum %2 !=0:
 max_length = max(max_length, right - left +1)
 left = right # If the current sum is even, reset it to zero and move the left pointer elif current_sum %2 ==0:
 current_sum =0 left = right # Move the right pointer right +=1 return max_length

**注释**

* 我们使用了两个指针，分别从左到右和从右到左扫描数组。
* 当我们遇到一个元素时，我们将其添加到当前和中。如果当前和为奇数，我们更新最大长度并移动两个指针。
* 如果当前和为偶数，我们重置它为零并移动左指针。

**第四题：**

**题目描述**

给定一个整数数组，要求找出其中最长的连续子序列，其元素之差绝对值不超过3。

**示例**

输入：[1,2,3,4,5]

输出：5（因为 [1,2,3,4,5] 是最长的连续子序列，其元素之差绝对值不超过3）

**代码实现**

def longest_abs_diff_not_exceed_3_subarray(nums):
 """
 Finds the longest continuous subarray with an absolute difference not exceeding3.

 Args:
 nums (list): A list of integers.

 Returns:
 int: The length of the longest continuous subarray with an absolute difference not exceeding3.
 """
 max_length =0 current_sum =0 # Initialize two pointers, one at the start and one at the end of the array left =0 right =0 while right < len(nums):
 # Add the element at the right pointer to the current sum current_sum += nums[right]

 # If the absolute difference between the current and previous elements is not exceeding3,
 # update the max length and move both pointers if abs(nums[right] - nums[left]) <=3:
 max_length = max(max_length, right - left +1)
 left = right # Otherwise, reset the current sum to zero and move the left pointer else:
 current_sum =0 left = right # Move the right pointer right +=1 return max_length

**注释**

* 我们使用了两个指针，分别从左到右和从右到左扫描数组。
* 当我们遇到一个元素时，我们将其添加到当前和中。如果当前和与前一个元素之间的绝对差值不超过3，我们更新最大长度并移动两个指针。
* 如果当前和与前一个元素之间的绝对差值大于3，我们重置它为零并移动左指针。

**第五题：**

**题目描述**

给定一个整数数组，要求找出其中最长的连续子序列，其元素之差绝对值不超过2。

**示例**

输入：[1,2,3,4,5]

输出：5（因为 [1,2,3,4,5] 是最长的连续子序列，其元素之差绝对值不超过2）

**代码实现**

def longest_abs_diff_not_exceed_2_subarray(nums):
 """
 Finds the longest continuous subarray with an absolute difference not exceeding2.

 Args:
 nums (list): A list of integers.

 Returns:
 int: The length of the longest continuous subarray with an absolute difference not exceeding2.
 """
 max_length =0 current_sum =0 # Initialize two pointers, one at the start and one at the end of the array left =0 right =0