LeetCode #830. Positions of Large Groups

题目

830. Positions of Large Groups

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解题方法

遍历数组，维护当前字符curletter，当前字符的个数curlettercount和返回值rat，如果当前字符不是curletter，就看看curlettercount是多少，超过3的话就把起始和终止位置加入到返回值rat里，最后还需要看看结尾处是不是一个连续的字符组。
时间复杂度：O(n)
空间复杂度：O(1)

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代码

class Solution:
    def largeGroupPositions(self, s: str) -> List[List[int]]:
        curletter = s[0]
        curlettercount = 1
        rat = []
        for i in range(1, len(s)):
            if s[i] == curletter:
                curlettercount += 1
            else:
                if curlettercount >= 3:
                    rat.append([i-curlettercount, i-1])
                curletter = s[i]
                curlettercount = 1
        else:
            if curlettercount >= 3:
                rat.append([i+1-curlettercount, len(s)-1])
        return rat