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<string> &dict) {        if(dict.empty() || start == end)            return 0;        queue<string>strq;        queue<int>depq;        strq.push(start);        depq.push(1);        string cur, nxt;        int depth;        unordered_set<string> 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;    }};