Leetcode_word-ladder(c++ version)

地址：http://oj.leetcode.com/problems/word-ladder/

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.

思路：1. 广度优先遍历搜索到的一定是最短路径(广搜是dijstra的特例).

             2. 一次广搜到结果后即可返回

设置visited防止回搜

参考代码：

class Solution {
public:
    int ladderLength(string start, string end, unordered_set &dict) {
        if(dict.empty() || start == end)
            return 0;
        queuestrq;
        queuedepq;
        strq.push(start);
        depq.push(1);
        string cur, nxt;
        int depth;
        unordered_set visited;
        while(!strq.empty())
        {
            nxt = cur = strq.front();
            strq.pop();
            depth = depq.front();
            depq.pop();
            if(cur == end)
                return depth;
            for(int i = 0; i < cur.length(); ++i)
            {
                for(char ch = 'a'; ch<='z'; ++ch)
                {
                    if(ch!=cur[i])
                    {
                        nxt[i] = ch;
                        if(dict.find(nxt)!=dict.end() && visited.find(nxt)==visited.end())
                        {
                            visited.insert(nxt);
                            strq.push(nxt);
                            depq.push(depth+1);
                        }
                        nxt = cur;
                    }
                }
            }
        }
        return 0;
    }
};