【LeetCode】438.找到字符串中所有字母异位词

题目：
438.找到字符串中所有字母异位词

思路：

所谓异位词，就是和原来的词在数量和字母类型上一致。可以用字典判断。
虽然滑动窗口可以做，但是双指针保证扫描一遍，保持窗口大小是目标词大小。真香！

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        def str_2_dict(s):
            s_dict = {}
            for c in s:
                s_dict[c] = s_dict.get(c, 0) + 1
            return s_dict
        left = 0
        right = len(p)
        ans = []
        windows = s[left:right]
        while right <= len(s):
            windows = s[left:right]
            # print(windows)
            if str_2_dict(windows) == str_2_dict(p):
                ans.append(left)
            left += 1
            right += 1
        return ans

字符串转换成字典比较太慢了，时间超了！！

def str_2_dict(s):
            s_dict = {}
            for c in s:
                s_dict[c] = s_dict.get(c, 0) + 1
            return s_dict

用26位的数组表示26个字母，相应索引下的数字表示出现的次数！！

alph_hash = [0]*26

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        def alph_hash(s):
            table = [0]*26
            for c in s:
                table[ord(c)-97] += 1
            return table
        left = 0
        right = len(p)-1
        target = alph_hash(p)
        ans = []
        windows = s[left:right+1]
        hash_windows = alph_hash(windows)
        while right < len(s):
            if hash_windows == target:
                ans.append(left)
            hash_windows[ord(s[left])-97] -= 1
            # 计算下一个hahs_windows
            left += 1
            right += 1
            if right >= len(s):
                break
            hash_windows[ord(s[right])-97] += 1
            print(hash_windows)
        return ans