Python-动态规划-最长公共子序列

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def lcs_length(x, y):
    m = len(x)
    n = len(y)
    c = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if x[i - 1] == y[j - 1]:  # i j 位置上的字符串匹配的时候，来自于左上方+1
                c[i][j] = c[i - 1][j - 1] + 1
            else:
                c[i][j] = max(c[i - 1][j], c[i][j - 1])
    # for _ in c:
    #     print(_)
    return c[m][n]  # 返回lcs长度


def lcs(x, y):
    m = len(x)
    n = len(y)
    c = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
    b = [[0 for _ in range(n + 1)] for _ in range(m + 1)]  # 1 左上方  2 上方  3 左方
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if x[i - 1] == y[j - 1]:  # i j 位置上的字符串匹配的时候，来自于左上方+1
                c[i][j] = c[i - 1][j - 1] + 1
                b[i][j] = 1
            elif c[i - 1][j] >= c[i][j - 1]:  # 来自上方
                c[i][j] = c[i - 1][j]
                b[i][j] = 2
            else:
                c[i][j] = c[i][j - 1]
                b[i][j] = 3
    return c[m][n], b  # 返回lcs长度，和lcs对应表


def lcs_trackback(x, y):
    # 从lcs对应表回溯，取值
    c, b = lcs(x, y)
    i = len(x)
    j = len(y)
    res = []
    while i > 0 and j > 0:
        if b[i][j] == 1:
            res.append(x[i - 1])
            i -= 1
            j -= 1
        elif b[i][j] == 2:
            i -= 1
        else:
            j -= 1
    return "".join(reversed(res))  # 因为回溯是倒序，所以将其倒序返值


print(lcs_trackback("ABCBDAB", "BDCABA"))