动态规划与找和相等的数组合

在动态规划中，我们会将问题分解成更小的子问题，然后解决这些子问题。在本文中，我们将探讨在一个数组中找到一个或几个数，这些数的和等于一个给定的数的算法。我们将使用动态规划来解决这个问题。我们还将讨论如何优化算法以及如何使用递归和备忘录来解决这个问题。

问题分解

第一步是将问题分解成更小的子问题。我们可以将问题分解为以下子问题：

给定一个数组A和一个数target，是否存在一个子数组，其中数的和等于target？
如果存在这样的子数组，我们可以找到它吗？
优化算法

我们现在可以优化算法来找到子数组。我们可以使用备忘录来存储子问题的解，这样我们就可以避免重复计算。我们还可以使用递归来解决问题。

def find_subarray(A, target):
    """
    Finds a subarray in A with a sum equal to target.

    Args:
        A: A list of numbers.
        target: The target sum.

    Returns:
        A list of numbers that sum to target, or None if no such subarray exists.
    """

    # Create a dictionary to store the subproblems we have already solved.
    memo = {}

    def helper(i, target):
        """
        Finds a subarray in A starting at index i with a sum equal to target.

        Args:
            i: The starting index of the subarray.
            target: The target sum.

        Returns:
            A list of numbers that sum to target, or None if no such subarray exists.
        """

        # If we have already solved this subproblem, return the cached result.
        if (i, target) in memo:
            return memo[(i, target)]

        # If we have reached the end of the array, there is no subarray that sums to target.
        if i == len(A):
            return None

        # Try including the current element in the subarray.
        subarray_with_current_element = helper(i + 1, target - A[i])
        if subarray_with_current_element is not None:
            return [A[i]] + subarray_with_current_element

        # Try excluding the current element from the subarray.
        subarray_without_current_element = helper(i + 1, target)
        if subarray_without_current_element is not None:
            return subarray_without_current_element

        # We have tried both including and excluding the current element, and neither worked.
        # So there is no subarray that sums to target.
        memo[(i, target)] = None
        return None

    # Call the helper function to find the subarray.
    return helper(0, target)

这个算法的时间复杂度是O(n^2)，其中n是数组的长度。我们可以使用一个动态规划算法来将时间复杂度降低到O(n^3)。

def find_subarray_dp(A, target):
    """
    Finds a subarray in A with a sum equal to target.

    Args:
        A: A list of numbers.
        target: The target sum.

    Returns:
        A list of numbers that sum to target, or None if no such subarray exists.
    """

    # Create a 2D array to store the subproblems we have already solved.
    dp = [[None for _ in range(target + 1)] for _ in range(len(A) + 1)]

    # Initialize the first row and column of the DP table.
    for i in range(len(A) + 1):
        dp[i][0] = []
    for j in range(target + 1):
        dp[0][j] = None

    # Fill in the rest of the DP table.
    for i in range(1, len(A) + 1):
        for j in range(1, target + 1):
            # Try including the current element in the subarray.
            if j - A[i - 1] >= 0 and dp[i - 1][j - A[i - 1]] is not None:
                dp[i][j] = [A[i - 1]] + dp[i - 1][j - A[i - 1]]

            # Try excluding the current element from the subarray.
            if dp[i - 1][j] is not None:
                dp[i][j] = dp[i - 1][j]

    # Return the solution to the original problem.
    return dp[len(A)][target]

这个算法的时间复杂度是O(n^3)，其中n是数组的长度。