LeetCode 809 - 情感丰富的文字

class Solution:
    def expressiveWords(self, s: str, words: List[str]) -> int:
        # 统计字符串中连续字符的信息, (字符, 连续个数)
        s_char_info = []  # type: List[List[int]]
        s_ch_set = set(s)

        for ch in s:
            if not s_char_info or s_char_info[-1][0] != ch:
                s_char_info.append([ch, 0])
            s_char_info[-1][1] += 1

        ans = 0
        for word in words:
            word_char_info = []  # type: List[List[int]]
            ok = True
            for ch in word:
                if ch not in s_ch_set:  # 如果出现s中不存在的字符, 提前终止
                    ok = False
                    break
                # 同理, 连续统计
                if not word_char_info or word_char_info[-1][0] != ch:
                    word_char_info.append([ch, 0])
                word_char_info[-1][1] += 1

            # 如果出现s不存在的字符(ok=False) 或者 不连续字符数不匹配(如 sasss和sa)
            if not ok or len(s_char_info) != len(word_char_info):
                continue

            for i in range(len(s_char_info)):
                # 不可扩展的情况：
                # s中某个字符连续次数不超过3, 但word中该位置的字符连续次数又不相等
                # word中某个位置的字符连续次数超过了s相应位置字符的次数(没办法缩)
                if (s_char_info[i][1] < 3 and s_char_info[i][1] != word_char_info[i][1] or
                        s_char_info[i][1] < word_char_info[i][1]):
                    ok = False
                    break
            if ok:
                ans += 1
        return ans