(遍历)python这道题怎么解？

Question

编写一个函数 traverse() 接受一个包含 n 个字符串的列表 tb，每个字符串包含 n 个小写字符 (a-z).
tb 表示 n 行 n 列的正方形 table。函数 returns 由以下过程生成的字符串 st，它遍历从左上角单元格开始到右下角单元格结束的网格。
在每一步中，程序都会水平向右移动或垂直向下移动，具体取决于两个单元格中的哪个单元格具有 \smaller" 字母（即按字母顺序出现较早的字母）。
然后将访问单元格中的字母添加到 st。如果平局，可以选择任何一个方向。
当到达 table 的右边缘或下边缘时，显然只有唯一的下一个单元格可以移动到。例如，traverse(["veus", "oxde", "oxlx", "hwuj"]) returns "veudexj"

所以 table 看起来像这样：

v  o  o  h
e  x  x  w
u  d  l  u
s  e  x  j

我是 python 的新手，我写了这段代码......但它只打印 "veuexj" 我会说问题出在这一行 if new_list[a - 1][b - 1] == new_list[a - 1][-2]: 中，它强制参数为跳过 'd' 字符。 #我也不知道怎么解决。

def traverse(tb_list):
    new_list = tb_list.copy()
    st = new_list[0][0]
    parameter = ''
    handler = 1
    for a in range(1, len(new_list)):
        
        for b in range(handler, len(new_list[a])):

            if new_list[a - 1][b - 1] == new_list[a - 1][-2]:
                parameter = new_list[a][b]

            elif new_list[a - 1][b - 1] > min(new_list[a - 1][b], new_list[a][b - 1]):
                parameter = min(new_list[a - 1][b], new_list[a][b - 1])

            elif new_list[a - 1][b - 1] < min(new_list[a - 1][b], new_list[a][b - 1]):
                parameter = min(new_list[a - 1][b], new_list[a][b - 1])

            st += parameter

        handler = b

    return st


print(traverse(["veus", "oxde", "oxlx", "hwuj"]))

Answer 1

你可以尝试这样的事情（解释添加为评论）：

def traverse(tb_list):
    lst = tb_list.copy() #Takes a copy of tb_list
    lst = list(zip(*[list(elem) for elem in lst])) #Transposes the list
    final = lst[0][0] #Sets final as the first element of the list
    index = [0,0] #Sets index to 0,0

    while True:
        x = index[0] #The x coordinate is the first element of the list
        y = index[1] #The y coordinate is the second element of the list

        if x == len(lst) - 1: #Checks if the program has reached the right most point of the table
            if y == len(list(zip(*lst))) - 1: #Checks if program has reached the bottommost point of the table
                return final #If both the conditions are True, it returns final (the final string)
            else:
                final += lst[x][y+1] #If the program has reached the right most corner, but not the bottommost, then the program moves one step down
                index = [x, y+1] #Sets the index to the new coordinates

        elif y == len(list(zip(*lst))) - 1: #Does the same thing in the previous if condition button in an opposite way (checks if program has reached bottommost corner first, rightmost corner next)
            if x == len(lst) - 1:
                return final
            else:
                final += lst[x + 1][y] #If the program has reached the bottommost corner, but not the rightmost, then the program moves one step right
                index = [x + 1, y]

        else: #If both conditions are false (rightmost and bottommost)
            if lst[x+1][y] < lst[x][y+1]: #Checks if right value is lesser than the bottom value
                final += lst[x+1][y] #If True, then it moves one step right and adds the alphabet at that index to final
                index = [x+1,y] #Sets the index to the new coords
            else: #If the previous if condition is False (bottom val > right val)
                final += lst[x][y+1] #Moves one step down and adds the alphabet at that index to final
                index = [x,y+1] #Sets the index to the new coords

lst = ["veus", "oxde", "oxlx", "hwuj"]
print(traverse(lst))

输出：

veudexj

我已将说明添加为评论，请花点时间阅读。如果您对代码的任何部分仍然不清楚，请随时问我。欢迎对 optimize/shorten 我的代码提出任何建议。

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