[leetcode 127]Word Ladder

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本文介绍了一种算法，用于寻找两个单词之间的最短转换路径，每次仅改变一个字母，并确保每一步骤中的单词都存在于给定的词典中。示例展示了从'hit'到'cog'的转换过程。

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

Only one letter can be changed at a time
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.

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        if (start.size() == 0 || end.size() == 0) {
            return 0;
        }
        if (start.size() != end.size()) {
            return 0;
        }
        vector<string> path;
        int level = 1;
        int count = 1;
        path.push_back(start);
        dict.erase(start);
        while (!path.empty()) {
            auto tmp = path.front();
            path.erase(path.begin());
            count--;
            for (int j = 0; j < tmp.size(); j++) {
                auto ms = tmp;
                for (auto i = 'a'; i < 'z'; i++) {
                    if (i == ms[j]) {
                        continue;
                    }
                    ms[j] = i;
                    if (ms == end) {
                        return level+1;
                    }
                    if (dict.find(ms) != dict.end()) {
                        path.push_back(ms);
                        dict.erase(ms);
                    }
                }
            }
            if (count == 0) {
                count = path.size();
                level++;
            }
            
        }
        return 0;
    }
};