[LeetCode-127] Word Ladder

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.

 

BPS。

 

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
        // bps
        queue<pair<string, int> > path;
        unordered_set<string> visited;
        path.push(make_pair(start, 1));
        visited.insert(start);
        string cur_string, str_left, str_right, str_conv;
        int cur_step;
        
        while (!path.empty()) {
            cur_string = path.front().first;
            cur_step = path.front().second;
            for (int i = 0; i < cur_string.length(); ++i) {
                str_left = cur_string.substr(0, i);
                str_right = cur_string.substr(i + 1, cur_string.length() - i - 1);
                for (char c = 'a'; c <= 'z'; ++c) {
                    str_conv = str_left;
                    str_conv.append(1, c);
                    str_conv.append(str_right);
                    
                    if (0 == str_conv.compare(end)) {
                        return cur_step + 1;
                    }
                    if ((dict.find(str_conv) != dict.end()) && (visited.find(str_conv) == visited.end())) {
                        visited.insert(str_conv);
                        path.push(make_pair(str_conv, cur_step + 1));
                    }
                }
            }
            path.pop();
        }
        return 0;
    }
};