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- # 方法一:遍历name为基础
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+ # 在两个字符串长度范围内,用两个指针指向两个字符串的首部,若两个元素相同则两指针各前进一步,否则typed的指针前进一步
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class Solution (object ):
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def isLongPressedName (self , name , typed ):
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i = j = 0
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- # 遍历name字符
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- while i < len (name ):
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- # name中字符未完,typed字符已经结束
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- if j >= len (typed ):
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- return False
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- elif name [i ] == typed [j ]:
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+ while i < len (name ) and j < len (typed ):
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+ if name [i ] == typed [j ]:
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i += 1
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j += 1
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- elif i >= 1 and typed [j ] == typed [j - 1 ]:
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- j += 1
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else :
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- return False
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- return True
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-
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- # 方法二:遍历typed为基础
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- class Solution (object ):
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- def isLongPressedName (self , name , typed ):
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- i , name_len = 0 , len (name )
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- # 遍历typed字符
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- for j in range (len (typed )):
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- # name字符未结束
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- if i < name_len and name [i ] == typed [j ]:
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- i += 1
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- elif j == 0 or typed [j ] != typed [j - 1 ]:
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- return False
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- # typed字符遍历完后,判断name字符是否也结束
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- return i == name_len
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+ j += 1
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+ return i == len (name )
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