Question

def ShortestDistance(maps,i,j,short,k): #i,j-->current position.
    x=0
    y=0
    for a in range(0, len(maps)):
        for b in range(0, len(maps[0])):
            if maps[a][b] == 2:
                x = a
                y = b

    if (i==x) and (j==y):
        short=min(short,k)
        return short

    maps[i][j]=3

    if within(i+1,j) and possible(maps,i+1,j):
        return ShortestDistance(maps,i+1,j,short,k+1)

    if within(i, j+1) and possible(maps,i, j+1):
        return ShortestDistance(maps,i,j+1,short,k+1)

    if within(i-1,j) and possible(maps,i-1,j):
        return ShortestDistance(maps,i-1,j,short,k+1)

    if within(i,j-1) and possible(maps,i,j-1):
        return ShortestDistance(maps,i,j-1,short,k+1)

    maps[i][j]=0

result=ShortestDistance(maps,0,0,longest,0)
if result is None:
    answer=-1
else:
    answer=result

我在这里要做的是使函数遍历函数中的所有if语句，而不管先前的if语句是否为true。例如，我希望代码中的第三个if语句也可以通过，即使第二个为true（在这种情况下，计算机似乎也没有经过第三个）。我的最终目标是为所有可能的情况找到k的最小值。我将如何使其成为可能？

Answer 1

使用这个：

def ShortestDistance(maps,i,j,short,k): #i,j-->current position.
    x=0
    y=0
    for a in range(0, len(maps)):
        for b in range(0, len(maps[0])):
            if maps[a][b] == 2:
                x = a
                y = b

    if (i==x) and (j==y):
        short=min(short,k)
        return short

    maps[i][j]=3
    resx = []
    if within(i+1,j) and possible(maps,i+1,j):
        resx.append(ShortestDistance(maps,i+1,j,short,k+1))

    if within(i, j+1) and possible(maps,i, j+1):
        resx.append(ShortestDistance(maps,i,j+1,short,k+1))

    if within(i-1,j) and possible(maps,i-1,j):
        resx.append(ShortestDistance(maps,i-1,j,short,k+1))

    if within(i,j-1) and possible(maps,i,j-1):
        resx.append(ShortestDistance(maps,i,j-1,short,k+1))

    maps[i][j]=0
    return min(resx)

result=ShortestDistance(maps,0,0,longest,0)
if result is None:
    answer=-1
else:
    answer=result

Answer 2

请注意return语句。一旦函数遇到一个，它就会退出。尝试这样的事情：

def ShortestDistance(maps,i,j,short,k):
    #...
    distances = []
    if within(i+1,j) and possible(maps,i+1,j):
        distances.append(ShortestDistance(maps,i+1,j,short,k+1))

    if within(i, j+1) and possible(maps,i, j+1):
        distances.append(ShortestDistance(maps,i,j+1,short,k+1))

    if within(i-1,j) and possible(maps,i-1,j):
        distances.append(ShortestDistance(maps,i-1,j,short,k+1))

    if within(i,j-1) and possible(maps,i,j-1):
        distances.append(ShortestDistance(maps,i,j-1,short,k+1))

    return min(distances)

是否可以遍历函数中的所有if语句？

2 个答案: