leetcode：查找

1. word ladder

题目：

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

Only one letter can be changed at a time

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 。所有的变化情况联系起来就是一棵树， 相当于从根结点start出发寻找特定的节点end。

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 = []

public ArrayList<String> generateParenthesis(int n) {
ArrayList<String> res = new ArrayList<String>();
String item = new String();

if (n<=0)
return res;

dfs(res,item,n,n);
return res;
}

// item保存上一步的临时结果，res最终结果； leftt,right指示搜索层级变化
public void dfs(ArrayList<String> res, String item, int left, int right){
if(left > right)//deal wiith ")("
return;

if (left == 0 && right == 0){
res.add(new String(item));
return;
}

if (left>0)
dfs(res,item+'(',left-1,right);
if (right>0)
dfs(res,item+')',left,right-1);
}

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