행맨(내 코딩 - list, for)

 

유명한 게임인 행맨을 만들어보았다.

 

책의 예제로 나와있는건데, 일부러 예제를 안보고 알고있는 지식과 검색만으로 코딩을 해보았다.

아마 책에 비하면 굉장히 괴랄한 코드일 것 같은데..

 

import random
import time
import sys
 
 
 
def getWord():
    return wordList[random.randint(0,len(wordList)-1)]
 
wordList = ['mouse', 'laptop', 'tablet', 'computer'] #문제에 나올 단어 지정
missingNum = 0 #단어를 틀린 횟수
missedLetters = '' #놓친 문자
selectedWord = getWord() #맞혀야 할 단어.
wordScreen = '_'*len(selectedWord) #단어를 맞히는 현황 표시
 
def printGraphic():
    #print out man.-------------------------------------------------
    print()
    print('-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-')
    print()
    print('H  A  N  G  M  A  N')
    print('     +------+')
    print('      l\tl')
 
    if missingNum >= 1:
        print('     O\tl')
        if missingNum == 1 :
            print('      \tl')
            print('      \tl')
        elif missingNum == 2 :
            print('      l\tl')
            print('      \tl')
        elif missingNum == 3 :
            print('    / l\tl')
            print('      \tl')
        elif missingNum == 4 :
            print('    / l＼\tl')
            print('      \tl')
        elif missingNum ==5 :
            print('    / l＼\tl')
            print('    /\tl')
        elif missingNum ==6 :
            print('    / l＼\tl')
            print('    /  ＼\tl')
    if missingNum == 0 :
        print('      \tl')
        print('      \tl')
        print('      \tl')
    
    print(' \tl')
    print('=========')
 
    #print out game state ------------------------------------------
    print('Missed letters : '+missedLetters)
    for i in wordScreen:
        print(i , end=' ')
    print()
    print('Guess a letter.')
    print(' >> Your Answer : ',end='')
 
 
def checkCorrect(letter):
    correct = False
 
    global selectedWord
    global wordScreen
 
    k=0
    temp=0
    for i in selectedWord:
        if i==letter:
            correct = True
 
            #i가 selectedWord에서 몇 번째 문자인지 알아내기
            while k<len(selectedWord):
                if i == selectedWord[k]:
                    temp=k
                    k+=1
                    break
                k+=1
 
            #wordScreen 내용 바꾸기
            temp_before = wordScreen[0:temp]
            temp_after = wordScreen[temp+1:len(selectedWord)]
            
            wordScreen = temp_before + letter +  temp_after
 
    return correct
            
def checkOverlap(letter):
    overlap = False
 
    for i in wordScreen:
        if letter == i:
            overlap = True
            break;
        
    for i in missedLetters:
        if letter == i:
            overlap = True
            break
 
    return overlap
 
def replayGuide():
    
    print('Do you want to play again? (yes / no)')
    while True:
        playAgain = input()
        if playAgain == 'yes':
            print()
            print()
            print()
            print()
            print()
            print('-=-=-=-=-=-=-=-=-=-=-=-=-')
            print(          'New game start')
            print('-=-=-=-=-=-=-=-=-=-=-=-=-')
            print()
            print()
            print()
            print()
            print()
            return playAgain
            break;
        elif playAgain=='no':
            print('Okay. Bye.')
            sys.exit(0)
        else :
            print('please type yes or no')
            continue
            
 
 
####################################################################
####################################################################
####################################################################
##
        #                            이 아래에서 위 함수들을 사용해 프로그램을 돌림
 
temp = 0
 
while True:
    temp = 0
    printGraphic()
    letter = input()
    letter = letter.lower()
    
    #철자가 하나인지 확인
    if len(letter) > 1:
        print('You should type just a letter. ( 1 letter )')
        continue
 
    #철자 중복여부 확인
    if checkOverlap(letter) == True :
        print('You have already guessed that letter. Choose again.')
        continue
 
    #주어진 단어안에 입력한 철자가 있는지 확인후 wordScreen 내용 바꾸기
    correct = checkCorrect(letter)
    if correct == False :
        missingNum+=1
        missedLetters += letter
 
    #정답을 모두 맞추었는지 확인
    for i in wordScreen:
        if i=='_':
            break
        
        #정답을 모두 맞춘경우
        if temp==len(wordScreen)-1 :    
            printGraphic()
            print()
            print()
            print('====================================================')
            print('You win!')
            if replayGuide()=='yes':
                missingNum = 0
                missedLetters = ''
                selectedWord = getWord()
                wordScreen = '_'*len(selectedWord)
                temp = 0
                break
            
        temp+=1
 
    #6회 이상 틀렸는지 확인 후 게임오버처리.
    if missingNum == 6:
        printGraphic()
        print()
        print()
        print('====================================================')
        print('Oh no... The man passed away.')
        print('You lose')
        if replayGuide()=='yes':
            missingNum = 0
            missedLetters = ''
            selectedWord = getWord()
            wordScreen = '_'*len(selectedWord)
            temp = 0
            continue
        
        
 
 
 
 
 
 
 
 
 
 
    
 

 

list : 다른 언어의 배열 처럼 사용할 수 있음. array와 같고 이름만 다른지, list가 자료구조할 때 그 list인지는 아직 모름

for : 신기하게 생김 https://wikidocs.net/22