LeetCode OJ

题目：

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.

Note:

• All words have the same length.
• All words contain only lowercase alphabetic characters.

解题思路：

　　广度优先搜索。另外设置一个标识位来避免死循环。

代码：

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        typedef pair<string, int> node;
        unordered_set<string> visited; //标识位，避免死循环
        queue<node> q;
        q.push(make_pair(start, 1));
        while (!q.empty()) {
            node cur_node = q.front();
            q.pop();

            if (cur_node.first == end) return cur_node.second;

            for (int i = 0; i < cur_node.first.size(); i++) {
                string tmp = cur_node.first;
                for (char j = 'a'; j <= 'z'; j++) {
                    tmp[i] = j; //change one character
                    if (tmp == end) return cur_node.second + 1;

                    if (dict.count(tmp) && !visited.count(tmp)) {
                        q.push(make_pair(tmp, cur_node.second + 1));
                        visited.insert(tmp);
                    }
                }
            }
        }
        return 0;
    }
};