[LeetCode]127. Word Ladder 深入浅出讲解和代码示例

1、汇总概要

本题解题思路中用到了哈希、堆栈的知识点。

2、题目

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

• Return 0 if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

3、审题

已知条件：起始单词、目的单词、词典

动作：起始单词每次变化其中一个字母，经过N次变化后，变为目的单词

求：从起始单词变化为目的单词的最小路径（N最小）

4、解题思路

以题目中的example来讲解思路。

本题的基本思路可以用暴力破解法，用多叉树的思路，找到所有从起始单词到目的单词的路径，再比较取最短路径即可。

但该方法空间复杂度消耗太大（需保存所有的多叉树路径），舍弃。

=> 本题用堆栈、哈希的思路来求解。

Step1：从起始单词到目的单词，可达路径用堆栈来保存，每次求取一组可达路径；

Step2：在迭代求取可达路径时，因已遍历过的路径不会保存，而只保留最优的一组，所以需标记记录已经迭代过的路径，这里用hash+list来设计标记。

建立个hash表，将beginWord、endWord、wordList中所有出现的单词建立一张映射表，key=word，value是一存放所有与该word相差一个字符的单词数组，如下：

{

'hit': ['hot'], 'lot': ['hot', 'dot', 'log'], 'dog': ['dot', 'log', 'cog'], 'hot': ['dot', 'lot', 'hit'], 'cog': ['dog', 'log'], 'dot': ['hot', 'dog', 'lot'], 'log': ['dog', 'lot', 'cog']}

Step3：迭代过程如下图（反复往stack中push和pop的过程）

1) 将起始单词(hit)放入stack堆栈;

2) 从hash表中查找hit对应的value，顺序从该数组中选取，此处选取"hot"放入堆栈;

3) 循环调用2)，依次将"dot","dog"...等放入堆栈 （见上图中最左处stack中的值）

4) 这里说明下"lot"，lot对应的value数组为[‘hot’,‘dot’,‘log’]，而hot,dot,log都已在stack中，即lot后无单词可往stack中放，该路径结束(弹出lot)

5) 继续按照4)的规则，依次弹出lot，放入cog，后又弹出cog、弹出log，放入cog，一条可达路径完成，记录下该路径；

6) 继续迭代，止到所有路径都遍历完毕（结束条件为stack中只剩下底部单词hit时）

5、代码示例 - Python

import timeimport copyclass Solution:    def dist(self,word1,word2):        len1 = len(word1)        len2 = len(word2)        disNum = len1        if(len1 == len2):            for i in range(0,len1):                if (word1[i] == word2[i]):                    disNum = disNum - 1        else:            disNum = disNum + 1        return disNum    def listToDict(self,begin,wd):        #build hash + link from wordlist        wd.append(begin) #add begin into dict        lenw = len(wd)        wdict = {}        for i in range (0,lenw):            for j in range(0,lenw):                dres = self.dist(wd[i],wd[j])                if(dres == 1):                    if wdict.has_key(wd[i]):                        wdict[wd[i]].append(wd[j])                    else:                        wdict[wd[i]] = [wd[j]]                    return wdict        def ifLast(self,key,value,wdict):        valuel = wdict[key]        index = valuel.index(value)        if index == len(valuel)-1:            return 0        else:            return 1                def findLabbers(self,begin,end,wd):        wdict = self.listToDict(begin,wd)        max = len(wd) + 1        wstack = []        index = 0        wstack.append(begin)        top = 1 #top is the stack length        toppre = ""        res = [[]];        reslen = 0        #build stack        pre = ""        while wstack:            print "\n\n go while ---"            time.sleep(0.1)            '''            1. push to stack            2. loop to push until top == 1 or stack[top] = end            3. if top == 1                    pop item and go to step 2                if stack[top] = end                    save the res and pop item and go to step 2                if index + 1 > next.len                    end the loop            '''            if wstack[top-1] == end:                #get one res                if reslen == 0:                    res[reslen] = copy.deepcopy(wstack)                    reslen = reslen + 1                else:                    #find the min one                    min = len(res[0])                                        if min > top:                        while reslen > 1:                            res.pop()                            reslen = reslen - 1                        res[reslen-1] = copy.deepcopy(wstack)                     else:                        if min == top:                            res.append("")                            res[reslen] = copy.deepcopy(wstack)                            reslen = reslen + 1                        else:                            print "min<top: the result > exist res, not put into res list"                 #loop to find another res                pre = wstack.pop()                 top = top - 1                if wstack:                    pre = wstack.pop()                    top = top - 1 #when get res, pop twice                            #if checked all the res, end the loop            if top == 1:                key = wstack[top-1]                if pre != "":                    if self.ifLast(key,pre,wdict):                        break #end the all loop                        # build the stack            if wdict.has_key(wstack[top-1]):                valuel = wdict[wstack[top-1]]                #get the pre index, and loop from it                               if pre == "" or pre not in valuel: #need remove duplicate here                    index = 0;                else:                    index = valuel.index(pre)                    index = index + 1                #loop to get the item, which not in stack                while(index <= len(valuel)-1):                    if (valuel[index] in wstack):                        print "go 7 ---"                        index = index + 1;                        print "index, len(valuel)-1",index, len(valuel)-1                    else:                        break                #check whether have item to put into stack                if index <= len(valuel)-1:                       #have item to put into stack                             lenws = len(wstack)                    #get the right item into stack                    #the stack length shouldn't to be too long                    if (lenws < max):                        wstack.append(valuel[index])                    else:                        #if not the right item, pop to reloop                        pre = wstack.pop()                        top = top - 1                        continue                else:                    #no item  to put into stack, pop to reloop                    pre = wstack.pop()                    top = top - 1                    continue                top = top + 1        return res    if __name__ == "__main__":    wd = ["hot","dot","dog","lot","log","cog"]    begin = "hit"    end = "log"    st = Solution()    res = st.findLabbers(begin,end,wd)    print "\n\nget result ==============\n",res

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