Word Ladder leetcode java

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Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

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.

题解：

 这道题是套用BFS同时也利用BFS能寻找最短路径的特性来解决问题。

 把每个单词作为一个node进行BFS。当取得当前字符串时，对他的每一位字符进行从a~z的替换，如果在字典里面，就入队，并将下层count++，并且为了避免环路，需把在字典里检测到的单词从字典里删除。这样对于当前字符串的每一位字符安装a~z替换后，在queue中的单词就作为下一层需要遍历的单词了。

 正因为BFS能够把一层所有可能性都遍历了，所以就保证了一旦找到一个单词equals（end），那么return的路径肯定是最短的。

 像给的例子 start = hit，end = cog，dict = [hot, dot, dog, lot, log]

 按照上述解题思路的走法就是：

  level = 1    hit   dict = [hot, dot, dog, lot, log]

         ait bit cit ... xit yit zit ，  hat hbt hct ... hot ... hxt hyt hzt ，  hia hib hic ... hix hiy hiz

  level = 2    hot  dict = [dot, dog, lot, log]

         aot bot cot dot ...  lot ... xot yot zot，hat hbt hct ... hxt hyt hzt，hoa hob hoc ... hox hoy hoz

  level = 3    dot lot  dict = [dog log]

         aot bot ... yot zot，dat dbt ...dyt dzt，doa dob ... dog .. doy doz，

         aot bot ... yot zot，lat lbt ... lyt lzt，loa lob ... log... loy loz

  level = 4   dog log dict = []

         aog bog cog

  level = 5   cog  dict = []

 代码如下：

Word Ladder leetcode java,布布扣,bubuko.com

Word Ladder leetcode java

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原文地址：http://www.cnblogs.com/springfor/p/3893499.html