leetcode -- Word Ladder

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 [问题描述]

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.

[解题思路]

BFS经典场景

int Solution::ladderLength(std::string start, std::string end, std::tr1::unordered_set<std::string> &dict)
{
    std::queue<std::pair<std::string, int> > bfs;
    std::tr1::unordered_set<std::string> path;
    path.insert(start);
    bfs.push(std::make_pair(start, 2));
    while(!bfs.empty()){
        std::string tmp = bfs.front().first;
        int num = bfs.front().second;
        for (int i = 0; i < tmp.length(); i ++){
            tmp = bfs.front().first;
            for(char ti = 'a'; ti <= 'z'; ti ++){
                tmp[i] = ti;
                if (tmp == end)
                    return num;
                if (dict.find(tmp) != dict.end() && path.find(tmp) == path.end()){
                    bfs.push(std::make_pair(tmp, num + 1));
                    path.insert(tmp);
                }
            }
        }
        bfs.pop();
    }
    return 0;
}