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python-3.x - 动态规划，创建备忘录表最长稳定子序列

问题描述

我已经研究了很长一段时间的动态编程问题并且被卡住了，所以非常感谢任何帮助。

这是我能够通过测试的问题的第一部分。 在此处输入图像描述

def lssLength(a, i, j):
    aj = a[j] if 0 <= j < len(a) else None
    # Implement the recurrence below. Use recursive calls back to lssLength
    assert 0 <= i <= len(a)
    if i >= len(a) or j >= len(a):
        return 0
    if aj and abs(a[i] - a[j]) > 1:
        return lssLength(a, i+1, j)
    if aj is None or (abs(a[i] - a[j]) <= 1 and i != j):
        return max(lssLength(a, i+1, j), lssLength(a, i+1, i) + 1)
    else:
        return lssLength(a, i+1, j)

这是我针对第一个问题的测试用例：

    def test_lss_length(self):
        # test 1
        n1 = lssLength([1, 4, 2, -2, 0, -1, 2, 3], 0, -1)
        print(n1)
        self.assertEqual(4, n1)
        # test 2
        n2 = lssLength([1, 2, 3, 4, 0, 1, -1, -2, -3, -4, 5, -5, -6], 0, -1)
        print(n2)
        self.assertEqual(8, n2)
        # test 3
        n3 = lssLength([0, 2, 4, 6, 8, 10, 12], 0, -1)
        print(n3)
        self.assertEqual(1, n3)
        # test 4
        n4 = lssLength(
            [4, 8, 7, 5, 3, 2, 5, 6, 7, 1, 3, -1, 0, -2, -3, 0, 1, 2, 1, 3, 1, 0, -1, 2, 4, 5, 0, 2, -3, -9, -4, -2, -3,
             -1], 0, -1)
        print(n4)
        self.assertEqual(14, n4)

现在我需要采用递归解决方案并将其转换为动态编程，这就是我卡住的地方。我正在使用与以前相同的测试，但测试失败了。在此处输入图像描述

def memoizeLSS(a):
    T = {}  # Initialize the memo table to empty dictionary
    # Now populate the entries for the base case
    # Now fill out the table : figure out the two nested for loops
    # It is important to also figure out the order in which you iterate the indices i and j
    # Use the recurrence structure itself as a guide: see for instance that T[(i,j)] will depend on T[(i+1, j)]
    n = len(a)
    for i in range(0, n+1):
        for j in range(-1, n+1):
            T[(i, j)] = 0
    for i in range(n-1, -1, -1):
        for j in range(n-1, -1, -1):
            if abs(a[i] - a[j]) > 1:
                try:
                    T[(i, j)] = max(0, T[(i, j+1)], T[(i+1, j)])
                except Exception:
                    T[(i, j)] = 0
            elif abs(a[i] - a[j]) <= 1 and i != j:
                T[(i, j)] = T[(i+1, j+1)] + 1
            else:
                T[(i, j)] = max(0, T[(i+1, j+1)])
    for i in range(n-2, -1, -1):
        T[(i, -1)] = max(T[(i+1, -1)], T[(i+1, 0)], T[(i, 0)], 0)
    return T

如果您已阅读所有这些内容，非常感谢您。我知道这很多，非常感谢您的时间。非常感谢任何指向阅读材料等的指针。如果需要更多详细信息，请告诉我。谢谢。

标签： python-3.xdynamic-programmingmemoization

解决方案

您的解决方案仅适用于第一个测试用例。以下是修正版：

def memoizeLSS(a):
    T = {}  # Initialize the memo table to empty dictionary
    n = len(a)
    for j in range(-1, n):
        T[(n, j)] = 0 # i = n and j

    # Now populate the entries for the base case
    # Now fill out the table : figure out the two nested for loops
    # It is important to also figure out the order in which you iterate the indices i and j
    # Use the recurrence structure itself as a guide: see for instance that T[(i,j)] will depend on T[(i+1, j)]
    n = len(a)
    for i in range(0, n + 1):
        for j in range(-1, n + 1):
            T[(i, j)] = 0            

    for i in range(n-1, -1, -1):
        for j in range(n-1, -1, -1):
            aj = a[j] if 0 <= j < len(a) else None 
            if aj != None and abs(a[i] - a[j]) > 1:
                T[(i, j)] = T[(i+1, j)]
                
            elif aj == None or abs(a[i] - a[j]) <= 1:
                T[(i, j)] = max(T[(i+1, i)] + 1, T[(i + 1, j)])
    for i in range(n-2, -1, -1):
        T[(i, -1)] = max(T[(i+1, -1)], T[(i+1, 0)], T[(i, 0)], 0)

    return T

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