2012-03-21 110 views
0

我遇到一个程序有问题,程序需要一个单词,并且一次更改一个字母,将该单词转换为目标单词。虽然,请记住,根据我给出的单词字典,转换后的单词必须是合法的单词。递归与列表

我很难弄清楚如何使它递归。该程序对其必须采取的步骤数量有限制。

编辑:我不能让持有人名单全球。

到目前为止我的代码:

def changeling(word,target,steps): 
    holderlist=[] 
    i=0 
    if steps<0 and word!=target: 
     return None 

    if steps!=-1: 
     for items in wordList: 
      if len(items)==len(word): 
       i=0 

       if items!=word: 
        for length in items: 

         if i==1: 
          if items[1]==target[1] and items[0]==word[0] and items[2:]==word[2:]: 
           if items==target: 
            print "Target Achieved" 
            holder.list.append(target) 
           holderlist.append(items) 
           changeling(items,target,steps-1) 

         elif i>0 and i<len(word)-1 and i!=1: 
          if items[i]==target[i] and items[0:i]==word[0:i] and items[i+1:]==word[i+1:]: 
           if items==target: 
            print "Target Achieved" 
           holderlist.append(items) 
           changeling(items,target,steps-1) 

         elif i==0: 
          if items[0]==target[0] and items[1:]==word[1:]: 
           if items==target: 
            print "Target Achieved" 
           holderlist.append(items) 
           changeling(items,target,steps-1) 

         elif i==len(word)-1: 
          if items[len(word)-1]==target[len(word)-1] and items[0:len(word)-1]==word[0:len(word)-1]: 
           if items==target: 
            print "Target Achieved" 
           holderlist.append(items) 
           changeling(items,target,steps-1) 
         else: 
          changeling(None,None,steps-1) 

         i+=1 

    return holderlist 

我最大的问题是,我的控股名单holderlist刷新每次我尝试让程序递归。

我可以解决它,如果我手动输入数据。这是我想要的程序来做:

changeling("find","lose",4) 
gives me: 
['fine','fond'] 
the program should then do: 
changeling('fine','lose',3) 
gives me: 
['line'] 
changeling('line','lose',2) 
gives me: 
['lone'] 
changeling('lone','lose',1) 
gives me: 
['lose'] 
Target Achieved 
+5

现在已经晚了这里,而且我不打算现在看这个节目。但事实上,你已经将它嵌入了九层深,这是一个强有力的指标,表明存在根本性错误。 – 2012-03-21 21:20:01

+0

恩,感谢您的意见。 – Unknown 2012-03-21 21:23:56

+1

就像一个想法:做'如果步骤!= -1:<长缩进块>',而不是'如果步骤== -1:返回holderlist',然后放置之前的缩进块(保存一个缩进) 。类似的,不是'如果项目!=单词:<长缩进块>',做'if items == word:break';使用这种策略,您可以轻松减少到三个/四个嵌套层次,让您实际可视化程序的逻辑流程。另外,是不是'holder.list.append(target)'一个错字? – 2012-03-21 21:33:48

回答

1

可能像

def distx(w1,w2): 
    if len(w1) != len(w2):return 100000 
    score=0 
    for i in range(len(w1)): 
     score += int(w1[i] != w2[i]) 
    return score 


word_list = ["fine","fond","line","lose","lone"] 

def changeling(guess,target,steps): 
    my_steps = [] 
    print "Guess:",guess 
    if target == guess:return [guess] 
    try:word_list.remove(guess) 
    except:pass 
    my_steps.append(guess) 
    if target != guess and steps >= 0: 
     this_step = [] 
     one_step_away = [w for w in word_list if distx(guess,w) == 1] 
     for k in one_step_away: 
      print "  %s->"%guess,k 
      this_step.append(changeling(k,target,steps-1)) 
    my_steps.append(this_step) 
    return my_steps 
tmp = changeling("find","lose",4) 
print tmp