题目内容

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']

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