Question

这是代码中断问题的解决方案，我没有找到解决方案。

间谍游戏：编写一个函数，该函数接受一个整数列表，如果顺序包含007，则返回True

def spy_game(nums):

   code = [0,0,7,'x']

   for num in nums:
       if num == code[0]:
           code.pop(0)   # code.remove(num) also works

   return len(code) == 1

这个for循环如何特别检查索引0，但如何检查每个索引号？

Answer 1

由于您执行了code.pop(0)，因此将从列表code中删除索引0处的元素。当您使用[0, 0, 7]调用函数时，将进行以下循环迭代：

nums = [0, 0, 7]
code = [0, 0, 7, 'x']

# First iteration
num = 0 (first element of nums)
num == code[0] is true
code[0] is removed
code = [0, 7, 'x']

# Second iteration
num = 0 (second element of nums)
num == code[0] is true
code[0] is removed
code = [7, 'x']

# Third iteration
num = 7 (third element of nums)
num == code[0] is true
code[0] is removed
code = ['x']

# End of the loop
All elements of nums have been iterated, so the loop is over

code = ['x']
len(code) == 1 is true
True is returned

Answer 2

def spy_game(nums):
    mylist =  list(enumerate(nums))
    myupdatedlist = []

    for a,b in mylist:
        if b ==0 or b ==7:
            myupdatedlist.append(tuple((a,b)))
    if myupdatedlist[0][1] == 7:
        return False
    else:
        return True
    pass

Answer 3

我是这样做的：

def myfunc(a):
    for c, b in enumerate(a):
        if b == 0 and a[c+1] == 0 and a[c+2] == 7:
            return True
    return False

myfunc([0, 2, 7])

输出

False

即使它特别指出索引0，for循环如何检查每个索引号？

3 个答案: