9/9 每日一題(回傳連續出現3次以上的元素開字串中開始到結束的索引)

 In a string s of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like s = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z", and "yy".

A group is identified by an interval [start, end], where start and end denote the start and end indices (inclusive) of the group. In the above example, "xxxx" has the interval [3,6].

A group is considered large if it has 3 or more characters.

Return the intervals of every large group sorted in increasing order by start index.

 

Example 1:

Input: s = "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the only large group with start index 3 and end index 6.

Example 2:

Input: s = "abc"
Output: []
Explanation: We have groups "a", "b", and "c", none of which are large groups.

Example 3:

Input: s = "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]
Explanation: The large groups are "ddd", "eeee", and "bbb".

 

Constraints:

• 1 <= s.length <= 1000
• s contains lowercase English letters only.

class Solution:

    def largeGroupPositions(self, s: str) -> List[List[int]]:

        s_dict=Counter(s)

        large=[]

        for k, v in s_dict.items():

            if v>=3:

                large.append(k)

        n,c=0,0

        res=[]

        start,end=None,None

        while n< len(s):

            if c>0 and s[n] != s[n-1]:

                #初始化

                c=0

                start,end=None,None

            if s[n] in large:

                c+=1 #開始計算是否超過三

                if not start and start !=0: #檢查start是否存在,避免從0開始被洗掉

                    start=n

                if c>=3 and n==len(s)-1:

                    res.append([start,n])

                elif c>=3 and s[n+1] != s[n]:

                    end=n

                    res.append([start,end])

            n+=1

        return res

                   

-----------------------------------

class Solution:
    def largeGroupPositions(self, s: str) -> List[List[int]]:
        ans = []
        i = 0

        while i < len(s):
            char = s[i]
            start = i
            current = i
            while current < len(s) and s[current] == char:
                current += 1
            if current - start >= 3:
                ans.append([start, current-1])
            i = current
        
        return ans