题解 | #字符串合并处理#
字符串合并处理
https://www.nowcoder.com/practice/d3d8e23870584782b3dd48f26cb39c8f
while True:
try:
str1, str2 = input().split()
combine = str1 + str2
# 排序
combinels = []
even = []
odd = []
nums = len(combine)
# 偶数排序
for i in range(0, nums, 2):
even.append(combine[i])
even = sorted(even)
# 奇数排序
for i in range(1, nums, 2):
odd.append(combine[i])
odd = sorted(odd)
# 得到排序后的列表
for i in range(0, nums):
if i%2 == 0:
combinels.append(even[i//2])
else:
combinels.append(odd[i//2])
temp = combinels
chars = ""
# 进制转换(只对A-F和0-9)
pattern = list("ABCDEFabcdef0123456789")
for i in temp:
if i in pattern:
char_16to10 = int(i,16)
char_10to2 = list(str(bin(char_16to10)[2:]))# 在这里,可能有的数字只有一位(比如1),我们需要添加前面的项000
char_10to2.reverse() # 翻转过后就是在后面添加000
if len(char_10to2)<4:
char_10to2.extend(["0"]*(4-len(char_10to2)))
char_10to2 = "".join(char_10to2)
char_2to16 = hex(int(char_10to2,2))[2:]
chars += char_2to16.upper() # A-F需要转换为大写
else:
chars += i
print("".join(chars))
except:
break
