华为OD机试真题-最长子字符串的长度（一）-2023年OD统一考试（C卷）---Python3--开源

题目：
在这里插入图片描述

考察内容：
思路转化：求出o字母出现偶次（o的索引）；环形–双倍字母；
方法1：循环变量双倍字母（保证环线），记录最大偶次，如果是，则记录left和并替换left位置，并添加子字符串；
方法2：直接求出双倍字母的索引，根据最大偶次，循环遍历索引求出子字符串。
代码：

"""
analyze:
环形如何实现：
偶次（0也包括）
aloloboalolobo-- alolob;loboal;boalol; alolob

input:
小写字母字符串

output:
int, o字母出现偶数次
eg:
alolobo
6

looxdolx
7
way:
把所有还有两个0的子字符串求出来

根据o的索引获取(先求出最大偶数）

"""

# s_temp = input()
#
# if "o" not in s_temp:
#     print(len(s_temp))
# else:
#     o_res = 0
#     temp_list = list()
#     o_num = s_temp.count("o")
#     # 取最大偶数
#     if o_num % 2 != 0:
#         o_num = o_num - 1
#     # print(o_num)
#     double_s = s_temp*2
#     left, right = 0, 0
#     o_index = 0
#     for i in range(len(double_s)):
#         if double_s[i] == "o":
#             o_res += 1
#             if o_res <= o_num:
#                 pass
#                 # right += 1
#             else:
#                 if double_s[left:i] not in temp_list:
#                     temp_list.append(double_s[left:i])
#                 left = double_s.find("o", o_index) + 1
#                 o_index = left
#                 # right += 1
#                 o_res = 2
#         elif o_res <= o_num:
#             pass
#             # right += 1
#     max_str = 0
#     for temp in temp_list:
#         max_str = max(0, len(temp))
#     print(temp_list, max_str)

# 优化
# s_temp = input()
#
# if "o" not in s_temp:
#     print(len(s_temp))
# else:
#     o_res = 0
#     temp_list = list()
#     o_num = s_temp.count("o")
#     # 取最大偶数
#     if o_num % 2 != 0:
#         o_num = o_num - 1
#     double_s = s_temp * 2
#     left, right = 0, 0
#     o_index = 0
#     for i in range(len(double_s)):
#         if double_s[i] == "o":
#             o_res += 1
#             if o_res > o_num:
#                 if double_s[left:i] not in temp_list:
#                     temp_list.append(double_s[left:i])
#                 # 获取o的索引
#                 left = double_s.find("o", o_index) + 1
#                 o_index = left
#                 o_res = 2
#     max_str = 0
#     for temp in temp_list:
#         max_str = max(0, len(temp))
#     print(temp_list, max_str)

# 方法根据o的索引获取

s_temp = input()

if "o" not in s_temp:
    print(len(s_temp))
else:
    o_res = 0
    temp_index_list = list()
    o_num = s_temp.count("o")
    o_index = 0
    double_s = s_temp * 2
    print(double_s)
    for i in range(o_num*2):
        index = double_s.find("o", o_index)
        o_index = index + 1
        temp_index_list.append(index)
    print(temp_index_list)
    # 取最大偶数
    if o_num % 2 != 0:
        o_num = o_num - 1
    res = list()
    left, right = 0, 0
    for i in range(len(temp_index_list)):
        if i+o_num < len(temp_index_list):
            res.append(double_s[left:temp_index_list[i+o_num]])
            # print(double_s[left:temp_index_list[i+2]])
            left = temp_index_list[i]+1
    print(res)