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830. Positions of Large Groups 较大分组的位置

转载 作者:大佬之路 更新时间:2024-01-31 14:16:46 25 4
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题目地址:https://leetcode.com/problems/positions-of-large-groups/description/

题目描述

Ina string S of lowercase letters, these letters form consecutive groups of the same character.

Forexample, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".

Call a group large if it has 3 or more characters. We would like the starting and ending positions of every large group.

Thefinal answer should be in lexicographic order.

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting  3 and ending positions 6.

Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.

Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

Note: 1 <= S.length <= 1000

题目大意

一个长字符串可以按照字符的连续出现,分组。每个组内都是一段连续的,字符相同的子字符串。

要求,长度不小于3的所有组的字符串起始和结束位置。

解题方法

直接暴力求解即可!从左到右遍历字符串,只要后面的字符和该组起始的字符相同,那么就是属于同一个组;否则,开辟一个新组,并且判断之前的这个组长度是否>=3,是的话进行保存。

第一遍提交没通过的原因是忘了判断,当字符串结束的时候也是一个组终止的标志。比如字符串"aaa"

class Solution:
    def largeGroupPositions(self, S):
        """
        :type S: str
        :rtype: List[List[int]]
        """
        groups = []
        before_index, before_char = 0, S[0]
        for i, s in enumerate(S):
            if s != before_char:
                if i - before_index >= 3:
                    groups.append([before_index, i - 1])
                before_index = i
                before_char = s
        if i - before_index >= 2:
            groups.append([before_index, i])
        return groups

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二刷的时候,对结尾的判断是添加了一个大写字符,这样的话在不打扰之前小写字符串的基础上,增加了结束符号。

class Solution:
    def largeGroupPositions(self, S):
        """
        :type S: str
        :rtype: List[List[int]]
        """
        S = S + "A"
        groups = []
        previndex, prevc = 0, ""
        for i, c in enumerate(S):
            if not prevc:
                prevc = c
                previndex = i
            elif prevc != c:
                if i - previndex >= 3:
                    groups.append([previndex, i - 1])
                previndex = i
                prevc = c
        return groups

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