求矩阵最少（或最多）路径和（python实现）

1.递归方式
求最短，最终状态即右下角

f(v, i, j) = min(f(v, i - 1, j), f(v, i, j - 1)) + v[i][j]
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最长只需将min改为max即可

import numpy as np
# i：行
# j：列
# v：矩阵


def f(v, i, j):
    if i == 0 and j == 0:
        return v[0][0]
    elif i == 0:
        return f(v, i, j - 1) + v[i][j]
    elif j == 0:
        return f(v, i - 1, j) + v[i][j]
    else:
        return min(f(v, i - 1, j), f(v, i, j - 1)) + v[i][j]


v = np.array([[1, 3, 5, 9], [8, 1, 3, 
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