word-ladder

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 length5.
Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

可以利用广度优先搜索寻找最短的路径,需要注意的是如果可以修改字典,那么就不需要另外的数组来存储是否访问过了.另外可以用last和nlast的方法记录层数

class Solution {public:int ladderLength(string start, string end, unordered_set<string> &dict) {    if (start == end)        return 0;    int count = 2;    queue<string> que;    que.push(start);    string last = start;    string nlast = start;    while (!que.empty())    {        string temp = que.front();        que.pop();        for (int i = 0; i<temp.size(); ++i)        {            char a = temp[i];            for (int k = 0; k<26; ++k)            {                temp[i] = k + 'a';                if (temp == end)                    return count;                if (dict.find(temp) != dict.end())                {                    que.push(temp);                    nlast = temp;                    dict.erase(temp);                }            }            temp[i] = a;    //恢复temp        }        if (temp == last)  //一层遍历完毕        {            ++count;            last = nlast;        }    }    return 0;}};