leetcode 127. Word Ladder、126. Word Ladder II

127. Word Ladder

这道题使用bfs来解决，每次将满足要求的变换单词加入队列中。

wordSet用来记录当前词典中的单词，做一个单词变换生成一个新单词，都需要判断这个单词是否在词典中，不在词典中就不能加入队列。

pathCnt用来记录遍历到的某一个词使用的次数，做一个单词变换生成一个新单词，都需要判断这个单词是否在pathCnt中，如果在，则说明之前已经达到过，这次不用再计算了，因为这次计算的path肯定比之前多。pathCnt相当于剪枝。

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set wordSet(wordList.begin(),wordList.end());
        if(!wordSet.count(endWord))
            return 0;
        unordered_map<string,int> pathCnt;
        queue<string> q;
        q.push(beginWord);
        pathCnt[beginWord] = 1;
        while(!q.empty()){
            string word = q.front();
            q.pop();
            for(int i = 0;i < word.size();i++){
                string newWord = word;
                for(char j = 'a';j <= 'z';j++){
                    newWord[i] = j;
                    if(newWord == endWord)
                        return pathCnt[word] + 1;
                    if(wordSet.count(newWord) && !pathCnt.count(newWord)){
                        q.push(newWord);
                        pathCnt[newWord] = pathCnt[word] + 1;
                    }
                }
            }
        }
        return 0;
    }
};

126. Word Ladder II

https://www.cnblogs.com/grandyang/p/4548184.html

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        vector<vector<string>> res;
        unordered_set<string> dict(wordList.begin(), wordList.end());
        vector<string> p{beginWord};
        queue<vector<string>> paths;
        paths.push(p);
        int level = 1, minLevel = INT_MAX;
        unordered_set<string> words;
        while (!paths.empty()) {
            auto t = paths.front(); paths.pop();
            if (t.size() > level) {
                for (string w : words) dict.erase(w);
                words.clear();
                level = t.size();
                if (level > minLevel) break;
            }
            string last = t.back();
            for (int i = 0; i < last.size(); ++i) {
                string newLast = last;
                for (char ch = 'a'; ch <= 'z'; ++ch) {
                    newLast[i] = ch;
                    if (!dict.count(newLast)) continue;
                    words.insert(newLast);
                    vector<string> nextPath = t;
                    nextPath.push_back(newLast);
                    if (newLast == endWord) {
                        res.push_back(nextPath);
                        minLevel = level;
                    } else paths.push(nextPath);
                }
            }
        }
        return res;
    }
};