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)