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  • leetcode 482. License Key Formatting

    You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

    Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

    Given a non-empty string S and a number K, format the string according to the rules described above.

    Example 1:

    Input: S = "5F3Z-2e-9-w", K = 4
    
    Output: "5F3Z-2E9W"
    
    Explanation: The string S has been split into two parts, each part has 4 characters.
    Note that the two extra dashes are not needed and can be removed.
    

    Example 2:

    Input: S = "2-5g-3-J", K = 2
    
    Output: "2-5G-3J"
    
    Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.
    

    超级无聊的一个题目,还做了好几次才对!!!其实就是先将前面几个异常的找出来,然后再规则处理。

    class Solution(object):
        def licenseKeyFormatting(self, S, K):
            """
            :type S: str
            :type K: int
            :rtype: str
            """
            arr = "".join(S.upper().split('-'))        
            left = len(arr)%K        
            grouped = [arr[:left]] if left>0 else []
            for i in xrange(left, len(arr), K):            
                grouped.append(arr[i:i+K])
            return '-'.join(grouped)
            
    

     OR:

     def licenseKeyFormatting(self, S, K):
            """
            :type S: str
            :type K: int
            :rtype: str
            """
            S = S.replace("-", "").upper()[::-1]
            return '-'.join([S[i:i+K] for i in range(0, len(S), K)])[::-1]
    
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  • 原文地址:https://www.cnblogs.com/bonelee/p/9114067.html
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