• 2022-06-04
    下面代码的输出结果是__________。letter=['A','B','C','D','D','D']foriinletter:ifi=='D':letter.remove(i)print(letter)
    A: ['A','B','C']
    B: ['A','B','C','D','D']
    C: ['A','B','C','D','D','D']
    D: ['A','B','C','D']