567. 字符串的排列【collections】【hash map】【滑动窗口】

之前不知道abc怎么和它的全排列做比较，这里使用的方法是通过字符映射到数字，再对数字进行求和。abc和cba的和是一样的。 

第一种：滑动窗口

For each window representing a substring of s2 of length len(s1), we want to check if the count of the window is equal to the count of s1. Here, the count of a string is the list of: [the number of a's it has, the number of b's,... , the number of z's.]

We can maintain the window by deleting the value of s2[i - len(s1)] when it gets larger than len(s1). After, we only need to check if it is equal to the target. Working with list values of [0, 1,..., 25] instead of 'a'-'z' makes it easier to count later.

def checkInclusion(self, s1, s2):
    A = [ord(x) - ord('a') for x in s1]
    B = [ord(x) - ord('a') for x in s2]
    
    target = [0] * 26
    for x in A:
        target[x] += 1
    
    window = [0] * 26
    for i, x in enumerate(B):
        window[x] += 1
        if i >= len(A):
            window[B[i - len(A)]] -= 1
        if window == target:
            return True
    return False

from collections import Counter
class Solution:
    def checkInclusion(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        d1, d2 = Counter(s1), Counter(s2[:len(s1)])
        for start in range(len(s1), len(s2)):
            if d1 == d2: return True
            d2[s2[start]] += 1
            d2[s2[start-len(s1)]] -= 1
            if d2[s2[start-len(s1)]] == 0: 
                del d2[s2[start-len(s1)]]
        return d1 == d2

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最后一种： 

虽然超过时间限制，但是这个思路很简单——用i定位长的字符长，然后取（i+短的字符串）这么多个长度，和短的字符串做比较。

        ctr1 = collections.Counter(s1)
        i = 0
        while i < len(s2) - len(s1) + 1:
            if s2[i] in ctr1:
                ctr2 = collections.Counter(s2[i: i+len(s1)])
                if ctr1 == ctr2: return True
            i += 1
        return False

link 

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补充知识：

        key = sum(map(hash, s1))
        now = sum(map(hash, s2[:ls1]))

 这是干嘛的？？(O_O)?

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