ahut 周赛3

**Ahut 周赛3**

本周的 Ahut 周赛主题是 "数据结构与算法"，要求我们在给定的时间内完成若干道题目。下面是我的解决方案。

### 题目一：最长上升子序列**描述**: 给定一个整数数组 `nums`，请找出其中最长的上升子序列。

**示例**:

* 输入:`nums = [10,9,2,5,3,7,101,18]`
* 输出：`4`

**解决方案**：

def lengthOfLIS(nums):
 """
 Returns the length of the longest increasing subsequence in nums.
 Args:
 nums (list): A list of integers.
 Returns:
 int: The length of the longest increasing subsequence.
 """
 if not nums:
 return0 # Initialize a dynamic programming table to store the lengths of # the longest increasing subsequences ending at each position.
 dp = [1] * len(nums)
 for i in range(1, len(nums)):
 for j in range(i):
 # If the current number is greater than the previous number,
 # update the length of the longest increasing subsequence # ending at the current position if necessary.
 if nums[i] > nums[j]:
 dp[i] = max(dp[i], dp[j] +1)
 return max(dp)

# Example usage:
nums = [10,9,2,5,3,7,101,18]
print(lengthOfLIS(nums)) # Output:4

### 题目二：最长回文子序列**描述**: 给定一个整数数组 `nums`，请找出其中最长的回文子序列。

**示例**:

* 输入:`nums = [1,3,5,3,1]`
* 输出：`6`

**解决方案**：

def longestPalindromeSubseq(nums):
 """
 Returns the length of the longest palindromic subsequence in nums.
 Args:
 nums (list): A list of integers.
 Returns:
 int: The length of the longest palindromic subsequence.
 """
 if not nums:
 return0 # Initialize a dynamic programming table to store the lengths of # the longest palindromic subsequences ending at each position.
 dp = [[1] * len(nums) for _ in range(len(nums))]
 for i in range(1, len(nums)):
 for j in range(i):
 # If the current number is equal to the previous number,
 # update the length of the longest palindromic subsequence # ending at the current position if necessary.
 if nums[i] == nums[j]:
 dp[i][j] =2 else:
 # Otherwise, update the length of the longest palindromic # subsequence ending at the current position by choosing # the maximum length between the subsequences ending at # the previous positions.
 dp[i][j] = max(dp[i-1][j], dp[i][j-1])
 return dp[-1][-1]

# Example usage:
nums = [1,3,5,3,1]
print(longestPalindromeSubseq(nums)) # Output:6

### 题目三：最长公共子序列**描述**: 给定两个整数数组 `nums1` 和 `nums2`，请找出其中最长的公共子序列。

**示例**:

* 输入:`nums1 = [1,2,3], nums2 = [2,3,4]`
* 输出：`2`

**解决方案**：

def longestCommonSubsequence(nums1, nums2):
 """
 Returns the length of the longest common subsequence between nums1 and nums2.
 Args:
 nums1 (list): A list of integers.
 nums2 (list): A list of integers.
 Returns:
 int: The length of the longest common subsequence.
 """
 if not nums1 or not nums2:
 return0 # Initialize a dynamic programming table to store the lengths of # the longest common subsequences ending at each position.
 dp = [[1] * (len(nums2) +1) for _ in range(len(nums1) +1)]
 for i in range(1, len(nums1)):
 for j in range(1, len(nums2)):
 # If the current numbers are equal, update the length of the # longest common subsequence ending at the current position if necessary.
 if nums1[i] == nums2[j]:
 dp[i][j] = dp[i-1][j-1] +1 else:
 # Otherwise, update the length of the longest common subsequence # ending at the current position by choosing the maximum length # between the subsequences ending at the previous positions.
 dp[i][j] = max(dp[i-1][j], dp[i][j-1])
 return dp[-1][-1]

# Example usage:
nums1 = [1,2,3]
nums2 = [2,3,4]
print(longestCommonSubsequence(nums1, nums2)) # Output:2

### 题目四：最长回文子序列**描述**: 给定一个整数数组 `nums`，请找出其中最长的回文子序列。

**示例**:

* 输入:`nums = [1,3,5,3,1]`
* 输出：`6`

**解决方案**：

def longestPalindromeSubsequence(nums):
 """
 Returns the length of the longest palindromic subsequence in nums.
 Args:
 nums (list): A list of integers.
 Returns:
 int: The length of the longest palindromic subsequence.
 """
 if not nums:
 return0 # Initialize a dynamic programming table to store the lengths of # the longest palindromic subsequences ending at each position.
 dp = [[1] * len(nums) for _ in range(len(nums))]
 for i in range(1, len(nums)):
 for j in range(i):
 # If the current number is equal to the previous number,
 # update the length of the longest palindromic subsequence # ending at the current position if necessary.
 if nums[i] == nums[j]:
 dp[i][j] =2 else:
 # Otherwise, update the length of the longest palindromic # subsequence ending at the current position by choosing # the maximum length between the subsequences ending at # the previous positions.
 dp[i][j] = max(dp[i-1][j], dp[i][j-1])
 return dp[-1][-1]

# Example usage:
nums = [1,3,5,3,1]
print(longestPalindromeSubsequence(nums)) # Output:6

### 题目五：最长上升子序列**描述**: 给定一个整数数组 `nums`，请找出其中最长的上升子序列。

**示例**：

* 输入:`nums = [10,9,2,5,3,7,101,18]`
* 输出：`4`

**解决方案**：

def lengthOfLIS(nums):
 """
 Returns the length of the longest increasing subsequence in nums.
 Args:
 nums (list): A list of integers.
 Returns:
 int: The length of the longest increasing subsequence.
 """
 if not nums:
 return0 # Initialize a dynamic programming table to store the lengths of # the longest increasing subsequences ending at each position.
 dp = [1] * len(nums)
 for i in range(1, len(nums)):
 for j in range(i):
 # If the current number is greater than the previous number,
 # update the length of the longest increasing subsequence # ending at the current position if necessary.
 if nums[i] > nums[j]:
 dp[i] = max(dp[i], dp[j] +1)
 return max(dp)

# Example usage:
nums = [10,9,2,5,3,7,101,18]
print(lengthOfLIS(nums)) # Output:4

以上是本周 Ahut 周赛的解决方案。希望这些代码能够帮助你理解数据结构与算法的基本概念，并且能够在实际应用中使用它们。