438. 找到字符串中所有字母异位词

题目链接	438. 找到字符串中所有字母异位词
思路	滑动窗口
题解链接	官方题解
关键点	顺序比较；判断的状态量可以依此变更时应当使用“滑动窗口”的方式进行更新
时间复杂度	\(O(m + (n-m)\times\sum)\)
空间复杂度	\(O(\sum)\)

代码实现：

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        s_len = len(s)
        p_len = len(p)
        
        if s_len < p_len:
            return []
        
        answer = []

        s_count = [0] * 26
        p_count = [0] * 26
        a_ord = ord('a')
        for i in range(p_len):
            s_count[ord(s[i]) - a_ord] += 1
            p_count[ord(p[i]) - a_ord] += 1
        
        if s_count == p_count:
            answer.append(0)
        
        for i in range(s_len - p_len):
            s_count[ord(s[i]) - a_ord] -= 1
            s_count[ord(s[i+p_len]) - a_ord] += 1
            if s_count == p_count:
                answer.append(i+1)
        return answer

python-优化版本的滑动窗口

思路：维护“差异量” 时间复杂度：由$O(n+(n-m)\times\sum)$下降至$O(n+m+\sum)$

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        s_len = len(s)
        p_len = len(p)
        
        if s_len < p_len:
            return []
        
        answer = []

        a_ord = ord('a')
        ch_diff_count = [0] * 26
        for i in range(p_len):
            ch_diff_count[ord(s[i]) - a_ord] += 1
            ch_diff_count[ord(p[i]) - a_ord] -= 1
        diff = sum([
            item != 0
            for item in ch_diff_count
        ])

        if diff == 0:
            answer.append(0)
        
        for i in range(s_len - p_len):
            index = ord(s[i]) - a_ord
            if ch_diff_count[index] == 1:
                diff -= 1
            elif ch_diff_count[index] == 0:
                diff += 1
            ch_diff_count[index] -= 1

            index = ord(s[i+p_len]) - a_ord
            if ch_diff_count[index] == -1:
                diff -= 1
            elif ch_diff_count[index] == 0:
                diff += 1
            ch_diff_count[index] += 1
            
            if diff == 0:
                answer.append(i+1)
        return answer