python - 这个 for 循环如何检查每个索引号，即使它特别指出索引 0？

Question

这是一个代码破坏问题的解决方案，但我没有得到解决方案。

SPY GAME: Write a function that takes in a list of integers and returns True if it contains 007 in order

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 循环如何检查每个索引号，尽管它特别指出索引 0？

Answer 1

由于执行了 code.pop(0)，您将从列表 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