leetcode 1002. Find Common Characters( Python )

描述

Given an array A of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.

You may return the answer in any order. Example 1:

Input: ["bella","label","roller"]
Output: ["e","l","l"]
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Example 2:

Input: ["cool","lock","cook"]
Output: ["c","o"]
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Note:

1 <= A.length <= 100
1 <= A[i].length <= 100
A[i][j] is a lowercase letter
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解析

这里的题意比较简单，只要在 A 中的每一个字符串有相同的数量的字符个数，就都放到结果列表中。

解答

class Solution(object):
def commonChars(self, A):
    """
    :type A: List[str]
    :rtype: List[str]
    """
    result = []
    func = map(lambda x:collections.Counter(x), A)          
    for i in range(26):
        c = chr(ord('a')+i)
        num = min([count[c] for count in func])
        if num:
            result+=[c]*num
    return result
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运行结果

Runtime: 76 ms, faster than 20.19% of Python online submissions for Find Common Characters.
Memory Usage: 12.1 MB, less than 6.07% of Python online submissions for Find Common Characters.	
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解析

上面是遍历了 a-z 的全部字母，可是字符串中可能只须要遍历几个便可，因此能够换一下思路，只找第一个字符串中所包含的字母，这样能够大大缩短期。

解答

class Solution(object):
def commonChars(self, A):
    """
    :type A: List[str]
    :rtype: List[str]
    """
    check = set(A[0])
    tmp = [[l] * min([a.count(l) for a in A]) for l in check]
    result = []
    for i in tmp:
        result+=i
    return result
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运行结果

Runtime: 24 ms, faster than 99.25% of Python online submissions for Find Common Characters.
Memory Usage: 11.9 MB, less than 61.10% of Python online submissions for Find Common Characters.	
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感谢支持 支付宝

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