[LeetCode]Word Ladder

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord toendWord, 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.

[LeetCode Source]

思路：这个题目要用到BFS，图论的解法。就是无权最短路径问题。

我们从一个点开始搜索，每次扩展一层，直到找到最后的单词。

我们采用两个队列，一个队列存储距离，另一个队列存储单词。

每次遍历到一个点后要将该点从Dict中删除掉，一种方法是遍历Dict找到相邻的点，

第二种是生成一个单词所有相邻单词然后搜寻。第一种方法会超时。

特别注意对Set操作时，如果采用erase会让原来的迭代器失败。比如：

for(auto it=wordDict.begin();it!=wordDict.end();++it){
if(next(now,*it)){
dis.push(distance+1);
q.push(*it);
wordDict.erase(it);
}
}

这样编译的确通过，但是会出现运行时错误，原因在于erase之后原来迭代次失效了，导致这个循环出错，一定要避免。

最后AC的代码。这个时间依然有点长还可以改进。

class Solution {
public:
int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
queue<string> q;
q.push(beginWord);
queue<int> dis;
dis.push(1);
wordDict.erase(beginWord);
while(!q.empty()){
string now = q.front();
int distance = dis.front();
if(now==endWord)
return distance;
q.pop();
dis.pop();
for(int i=0;i<now.size();++i){
for(char j='a';j<='z';++j){
char rec = now[i];
now[i] = j;
if(wordDict.find(now)!=wordDict.end())
{
q.push(now);
dis.push(distance+1);
wordDict.erase(now);
}
now[i] = rec;
}
}
}
return 0;
}
};