LeetCode —— Word Ladder

链接：http://leetcode.com/onlinejudge#question_127

原题：

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：求最短路径，我用的是BFS，如果直接找set中只差一个字母的单词会超时，但是

遍历字母a~z，然后用hash表的优势，就不会超时了。

代码：

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        dict.insert(start);
        dict.insert(end);
        
        dict.erase(start);
        queue<pair<string, int> > q;
        q.push(pair<string, int>(start, 1));
        
        while (!q.empty()) {
            pair<string, int> cur = q.front();
            if (cur.first == end)
                return cur.second;
            q.pop();
            for (int n=0; n<(cur.first).size(); n++) {
                string word = cur.first;
                int stop = word[n] - 'a';
                for (int i=(stop+1)%26; i!=stop; i=(i+1)%26) {
                    word[n] = 'a' + i;
                    if (dict.find(word) != dict.end()) {
                        q.push(pair<string, int>(word, cur.second+1));
                        dict.erase(word);
                    }
                }
            }
        }
        return 0;
    }
};