题解 | #字符串合并处理#

字符串合并处理

https://www.nowcoder.com/practice/d3d8e23870584782b3dd48f26cb39c8f

import sys

Aword = str(input())
Aword = Aword.split(" ")

word1 = Aword[0]
word2 = Aword[1]
word = ""

len1 = len(word1)
len2 = len(word2)
lenmin = min(len1,len2)
lennum = len1 + len2

# 合并

word = Aword[0] + Aword[1]


# 排序
reline1 = []
reline2 = []
for i in range(len(word)):
    if i % 2 == 0:
        reline1.append(word[i])
    else:
        reline2.append(word[i])

for i in range(len(reline1)):
    for j in range(0, len(reline1) - i - 1):
        if reline1[j] > reline1[j + 1]:
            reline1[j] , reline1[j+1] = reline1[j+1] , reline1[j]

for i in range(len(reline2)):
    for j in range(0, len(reline2) - i - 1):
        if reline2[j] > reline2[j + 1]:
            reline2[j] , reline2[j+1] = reline2[j+1] , reline2[j]

#再次合并,并且对0-f 转换
newword = ""
for i in range(len(word)):
    sourcekey = ""
    if i % 2 == 0:
        sourcekey = reline1[int(i / 2)]
    else:
        sourcekey = reline2[int(i / 2)]
    if sourcekey == "0":
        newword += "0"
    elif sourcekey == "1":
        newword += "8"
    elif sourcekey == "2":
        newword += "4"
    elif sourcekey == "3":
        newword += "C"
    elif sourcekey == "4":
        newword += "2"
    elif sourcekey == "5":
        newword += "A"
    elif sourcekey == "6":
        newword += "6"
    elif sourcekey == "7":
        newword += "E"
    elif sourcekey == "8":
        newword += "1"
    elif sourcekey == "9":
        newword += "9"
    elif sourcekey == "A" or sourcekey == "a":
        newword += "5"
    elif sourcekey == "B" or sourcekey == "b":
        newword += "D"
    elif sourcekey == "C" or sourcekey == "c":
        newword += "3"
    elif sourcekey == "D" or sourcekey == "d":
        newword += "B"
    elif sourcekey == "E" or sourcekey == "e":
        newword += "7"
    elif sourcekey == "F" or sourcekey == "f":
        newword += "F"
    else:
        newword += sourcekey

print(newword)

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