文巾解题 567. 字符串的排列

Posted 2021-09-04 UQI-LIUWJ

1 题目描述

 2 解题思路：滑动窗口：

因为不管排列怎么排，s1字符串中的各个字母的个数是不变的。所以我们可以先设定一个大小和s1一样的滑动窗口。然后对s1和滑动窗口扫到的s2部分进行排序。如果而这一样，就说明包含这样的排列

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        needs, window = defaultdict(int), defaultdict(int)
        left, right = 0, 0
        valid = 0
        for i in s1:
            needs[i] += 1

        while right < len(s2):
            c = s2[right]
            right += 1
            if c in needs:
                window[c] += 1
                if window[c] == needs[c]:
                    valid += 1
            print(left, right)
            while right - left >= len(s1):
                if valid == len(needs):
                    return True
                d = s2[left]
                left += 1
                if d in needs:
                    if window[d] == needs[d]:
                        valid -= 1
                    window[d] -= 1
        return False